Some Basic Concepts of Chemistry JEE JEE Main| JEE Main Important Questions

Q 1 :

A sample of $C a C O_{3}$ and $M g C O_{3}$ weighed 2.21 g is ignited to constant weight of 1.152 g. The composition of mixture is

(Given molar mass in g $m o l^{- 1}$ $C a C O_{3} : 100, M g C O_{3} : 84$ ) [2024]

$1.023 g C a C O_{3} + 1.023 g M g C O_{3}$
$1.187 g C a C O_{3} + 1.187 g M g C O_{3}$
$1.023 g C a C O_{3} + 1.187 g M g C O_{3}$
$1.187 g C a C O_{3} + 1.023 g M g C O_{3}$

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(4)

$Let mass of {CaCO}_{3} in the mixture i s x g, so mass of {MgCO}_{3} in the mixture is (2.21 - x) g .$

$Moles of {CaCO}_{3} (n_{{CaCO}_{3}}) = \frac{given mass}{molar mass} = \frac{x}{100} mol$

$Moles of {MgCO}_{3} (n_{{MgCO}_{3}}) = \frac{given mass}{molar mass} = \frac{2.21 - x}{84} mol$

When the mixture is heated carbon dioxide escapes out leaving solid mass containing CaO and MgO.

${CaCO}_{3} (s) \overset{Δ}{\to} C a O (s) + {CO}_{2} (g)$

${MgCO}_{3} (s) \overset{Δ}{\to} M g O (s) + {CO}_{2} (g)$

By stoichiometry:

$Moles of CaO (n_{CaO}) = n_{{CaCO}_{3}} = \frac{x}{100} mol$

$Moles of MgO (n_{MgO}) = n_{{MgCO}_{3}} = \frac{2.21 - x}{84} mol$

$Mass of CaO (W_{CaO}) = n_{CaO} \times molar mass = \frac{x}{100} \times 56 g$

$Mass of MgO (W_{MgO}) = n_{MgO} \times molar mass = \frac{2.21 - x}{84} \times 40 g$

$Mass of residue = W_{CaO} + W_{MgO}$

$= \frac{x}{100} \times 56 + \frac{2.21 - x}{84} \times 40 = 1.152$

$\frac{56 x}{100} + \frac{(88.4 - 40 x)}{84} = 1.152$

$\frac{56 \times 84 x + (88.4 - 40 x) 100}{8400} = 1.152$

$704 x + 8840 = 9676.8$

$704 x = 836.8$

$x = 1.187$

$Thus mass of {CaCO}_{3} in the mixture is 1.187 g and mass of {MgCO}_{3} in the mixture is (2.21 - 1.187) g = 1.023 g .$

Q 2 :

The number of moles of methane required to produce 11g $C O_{2}$ (g) after complete combustion is:

[Given molar mass of methane in g $m o l^{- 1}$ : 16] [2024]

0.75
0.5
0.35
0.25

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(4)

${CH}_{4} (g) + 2 O_{2} (g) \to {CO}_{2} (g) + 2 H_{2} O (l)$

$Moles of {CO}_{2} (n_{{CO}_{2}}) = \frac{Given mass}{Molar mass} = \frac{11 g}{44 g {mol}^{- 1}} = 0.25 mol$

By stoichiometry of reaction,

$Moles of {CH}_{4} (n_{{CH}_{4}}) = Moles of {CO}_{2} (n_{{CO}_{2}}) = 0.25 mol .$

Q 3 :

Combustion of glucose $(C_{6} H_{12} O_{6})$ produces $C O_{2}$ and water. The amount of oxygen (in g) required for the complete combustion of 900g of glucose is: [Molar mass of glucose in g $m o l^{- 1}$ = 180] [2024]

480
32
800
960

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(4)

$Moles of glucose (n_{C_{6} H_{12} O_{6}}) = \frac{Given mass}{Molar mass}$

$= \frac{900 g}{180 g {mol}^{- 1}} = 5 mol$

$C_{6} H_{12} O_{6} (s) + 6 O_{2} (g) \to 6 {CO}_{2} (g) + 6 H_{2} O (l)$

By stoichiometry:

$\frac{n_{O_{2}}}{6} = \frac{n_{C_{6} H_{12} O_{6}}}{1}$

$\frac{n_{O_{2}}}{6} = \frac{5}{1}$

$n_{O_{2}} = 30 mol$

$W_{O_{2}} = n_{O_{2}} \times M_{O_{2}} = 30 mol \times 32 g m o l^{- 1} = 960 g$

Q 4 :

Mass of methane required to produce 22 g of $C O_{2}$ after complete combustion is __________ g.

(Given Molar mass in g $m o l^{- 1}$ ,C = 12.0, H = 1.0, O = 16.0) [2024]

(8) Number of moles of carbon dioxide

$(n_{{CO}_{2}}) = \frac{Given mass}{Molar mass} = \frac{22 g}{44 g {mol}^{- 1}} = 0.5 mol$

Combustion reaction of methane is represented as:

${CH}_{4} + 2 O_{2} \to {CO}_{2} + 2 H_{2} O$

By stoichiometry: $\frac{n_{{CH}_{4}}}{1} = \frac{n_{{CO}_{2}}}{1}, n_{{CH}_{4}} = 0.5 mol$

Mass of methane = $n_{{CH}_{4}} \times M_{{CH}_{4}} = 0.5 mol \times 16 g m o l^{- 1} = 8 g$

Q 5 :

If 50 mL of 0.5 M oxalic acid is required to neutralize 25 mL of NaOH solution, the amount of NaOH in 50 mL of given NaOH solution is _________ g. [2024]

(4) $H_{2} C_{2} O_{4} + 2 NaOH \to {Na}_{2} C_{2} O_{4} + 2 H_{2} O$

By stoichiometry:

$\frac{n_{NaOH}}{2} = \frac{n_{H_{2} C_{2} O_{4}}}{1}$

$\frac{M_{NaOH} \times V_{NaOH}}{2} = \frac{M_{H_{2} C_{2} O_{4}} \times V_{H_{2} C_{2} O_{4}}}{1}$

$\frac{M_{NaOH} \times 25 mL}{2} = \frac{0.5 M \times 50 mL}{1}$

$M_{NaOH} = 2 M$

For 50 mL, 2M NaOH solution:

$n_{NaOH} = molarity (M) \times volume (V)$

$= 2 M \times 50 mL = 100 mmol = 0.1 mol$

$Mass of NaOH = moles of NaOH \times molar mass of NaOH$

$= 0.1 mol \times 40 g m o l^{- 1} = 4 g$

Q 6 :

Number of moles of methane required to produce 22g $C O_{2 (g)}$ after combustion is $x \times 10^{- 2}$ moles. The value of $x$ is _________ . [2024]

(50) $Number of moles of carbon dioxide (n_{{CO}_{2}})$

$= \frac{Given mass}{Molar mass} = \frac{22 g}{44 g {mol}^{- 1}} = 0.5 mol$

Combustion reaction of methane is represented as:

$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$

By stoichiometry:

$\frac{n_{C H_{4}}}{1} = \frac{n_{C O_{2}}}{1}, n_{C H_{4}} = 0.5 mol = 50 \times 10^{- 2} mol$

Q 7 :

Consider the following reaction:

$3 {PbCl}_{2} + 2 {({NH}_{4})}_{3} {PO}_{4} \to {Pb}_{3} {({PO}_{4})}_{2} + 6 {NH}_{4} Cl$

If 72 mmol of $P b C l_{2}$ is mixed with 50 mmol of ${(N H_{4})}_{3} P O_{4}$ , then the amount of $P b_{3} {(P O_{4})}_{2}$ formed is _______ mmol (nearest integer). [2024]

(24) $3 {PbCl}_{2} + 2 {({NH}_{4})}_{3} {PO}_{4} \to {Pb}_{3} {({PO}_{4})}_{2} + 6 {NH}_{4} Cl$

Finding limiting reagent:

$\frac{n_{{PbCl}_{2}}}{Stoichiometric coefficient of {PbCl}_{2}} = \frac{72}{3} = 24$

$\frac{n_{{({NH}_{4})}_{3} P O_{4}}}{Stoichiometric coefficient of {({NH}_{4})}_{3} P O_{4}} = \frac{50}{2} = 25$

As 24 < 25, $P b C l_{2}$ is limiting reagent. Amount of product i.e. $P b_{3} {(P O_{4})}_{2}$ formed depends upon amount of limiting reagent i.e. $P b C l_{2}$ :

By stoichiometry:

$\frac{n_{{Pb}_{3} {({PO}_{4})}_{2}}}{1} = \frac{n_{{PbCl}_{2}}}{3}$

$\frac{n_{{Pb}_{3} {({PO}_{4})}_{2}}}{1} = \frac{72 mmol}{3}$

$n_{{Pb}_{3} {({PO}_{4})}_{2}} = 24 mmol$

Q 8 :

10 mL of gaseous hydrocarbon on combustion gives 40 mL of $C O_{2} (g)$ and 50 mL of water vapour. Total number of carbon and hydrogen atoms in the hydrocarbon is ________ [2024]

(14) $C_{x} H_{y} + (x + \frac{y}{4}) O_{2} \to x {CO}_{2} + (\frac{y}{2}) H_{2} O$

By stoichiometry:

$\frac{V_{C_{x} H_{y}}}{1} = \frac{V_{{CO}_{2}}}{x} = \frac{V_{H_{2} O}}{(\frac{y}{2})}$

$\frac{10}{1} = \frac{40}{x} = \frac{50}{(\frac{y}{2})}$

$\Rightarrow x = 4, y = 10$

$The hydrocarbon is C_{4} H_{10} . Total number of carbon and hydrogen atoms in it is 14 .$

Q 9 :

9.3 g of aniline is subjected to reaction with excess of acetic anhydride to prepare acetanilide. The mass of acetanilide produced if the reaction is 100% completed is ______ $\times 10^{- 1}$ g.

(Given molar mass in g ${mol}^{- 1}$ N : 14, O : 16
C : 12, H : 1) [2024]

(135)

Moles of aniline $(n_{C_{6} H_{5} {NH}_{2}})$ $= \frac{Mass of aniline}{M_{C_{6} H_{5} {NH}_{2}}} = \frac{9.3 g}{93 g {mol}^{- 1}} = 0.1 mol$

As aniline is the limiting reagent, moles of acetanilide produced depend upon moles of aniline.

By stoichiometry:

moles of acetanilide = moles of aniline

moles of acetanilide $(n_{C_{6} H_{5} {NHCOCH}_{3}}) = 0.1 mol$

$Mass of acetanilide = n_{C_{6} H_{5} {NHCOCH}_{3}} \times M_{C_{6} H_{5} {NHCOCH}_{3}}$

$= 0.1 mol \times 135 g {mol}^{- 1}$

$= 13.5 g = 135 \times 10^{- 1} g$

Q 10 :

${CaCO}_{3} (s) + 2 HCl (aq) \to {CaCl}_{2} (aq) + {CO}_{2} (g) + H_{2} O (l)$

Consider the above reaction, what mass of ${CaCl}_{2}$ will be formed if 250 mL of 0.76 M HCl reacts with 1000 g of ${CaCO}_{3}$ ?

(Given: Molar mass of Ca, C, O, H and Cl are 40, 12, 16, 1 and 35.5 g ${mol}^{- 1}$ , respectively) [2025]

3.908 g
2.636 g
10.545 g
5.272 g

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(3)

Moles of ${CaCO}_{3}$ $= \frac{{Mass of CaCO}_{3}}{{Molar mass of CaCO}_{3}} = \frac{1000}{100} mol = 10 mol$

$Moles of HCl = Molarity \times Volume in L = 0.76 M×0.25 L = 0.19 mol$

${CaCO}_{3} + 2 HCl \to {CaCl}_{2} + {CO}_{2} + H_{2} O$

As per reaction stoichiometry, 20 mol of HCl are needed to react with 10 mol ${CaCO}_{3}$ , but only 0.19 mol of HCl is available. Thus, HCl is the limiting reagent. Moles of ${CaCl}_{2}$ formed depends upon moles of limiting reagent i.e. HCl.

As per reaction stoichiometry, ${Moles of CaCl}_{2} (n_{{CaCl}_{2}}) = \frac{n_{HCl}}{2} = \frac{0.19}{2} mol$

${Mass of CaCl}_{2} = {Moles of CaCl}_{2} \times {Molar mass of CaCl}_{2} = \frac{0.19}{2} \times 111 g = 10.545 g$

Q 11 :

0.1 M solution of $KI$ reacts with excess of $H_{2} {SO}_{4}$ and ${KIO}_{3}$ solutions. According to equation

$5 I^{-} + {IO}_{3}^{-} + 6 H^{+} \to 3 I_{2} + 3 H_{2} O$

Identify the correct statements:

(A) 200 mL of KI solution reacts with 0.004 mol of ${KIO}_{3}$
(B) 200 mL of KI solution reacts with 0.006 mol of $H_{2} {SO}_{4}$
(C) 0.5 L of KI solution produced 0.005 mol of $I_{2}$
(D) Equivalent weight of ${KIO}_{3}$ is equal to $(\frac{molecular weight}{5})$

Choose the correct answer from the options given below: [2025]

(A) and (B) only
(A) and (D) only
(B) and (C) only
(C) and (D) only

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(2)

(A) $Moles of KI = Molarity \times Volume in L = 0.1 M \times 0.2 L = 0.02 mol$

As per reaction stoichiometry:

$\frac{n_{{KIO}_{3}}}{1} = \frac{n_{KI}}{5}$

$\frac{n_{{KIO}_{3}}}{1} = \frac{0.02}{5}$

$n_{{KIO}_{3}} = 0.004 mol$

(B) $Moles of KI = Molarity \times Volume in L$

$=0.1 M×0.2 L = 0.02 mol$

As per reaction stoichiometry:

$\frac{n_{H^{+}}}{6} = \frac{n_{KI}}{5}$

$\frac{2 \times n_{H_{2} {SO}_{4}}}{6} = \frac{n_{KI}}{5}$

$\frac{2 \times n_{H_{2} {SO}_{4}}}{6} = \frac{0.02}{5}$

$n_{H_{2} {SO}_{4}} = 0.012 mol$

(C) $Moles of KI = Molarity \times Volume in L$

$= 0.1 M×0.5 L = 0.05 mol$

As per reaction stoichiometry:

$\frac{n_{I_{2}}}{3} = \frac{n_{KI}}{5}$

$\frac{n_{I_{2}}}{3} = \frac{0.5}{5}$

$n_{I_{2}} = 0.50 mol$

(D) $n {factor of KIO}_{3} = ({Change in O.N. of I from KIO}_{3} {to I}_{2}) \times {Number of I in KIO}_{3} = (5) \times 1 = 5$

${Equivalent weight of KIO}_{3} = \frac{{Molecular weight of KIO}_{3}}{n factor} = \frac{{Molecular weight of KIO}_{3}}{5}$

Q 12 :

Mass of magnesium required to produce 220 mL of hydrogen gas at STP on reaction with excess of dil. HCl is

Given: Molar mass of Mg is 24 g ${mol}^{- 1}$ . [2025]

235.7 g
0.24 mg
236 mg
2.444 g

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(3)

Moles of hydrogen gas $(n_{H_{2}})$

$= \frac{Volume of hydrogen at STP in mL}{22400} = \frac{220}{22400} mol$

$Mg + 2 HCl \to {MgCl}_{2} + H_{2}$

As per reaction stoichiometry, moles of Mg $(n_{Mg})$ = $n_{H_{2}}$

$= \frac{220}{22400} mol$

Mass of Mg $(W_{Mg})$

$= n_{Mg} \times M_{Mg}$

$= \frac{220}{22400} \times 24 g = 0.2357 g = 235.7 mg$

Q 13 :

When 81.0 g of aluminium is allowed to react with 128.0 g of oxygen gas, the mass of aluminium oxide produced in grams is _______. (Nearest integer)

Given:
Molar mass of Al is 27.0 g ${mol}^{- 1}$
Molar mass of O is 16.0 g ${mol}^{- 1}$ [2025]

(153)

Given moles of aluminium $= \frac{Given mass of Al}{Molar mass of Al} = \frac{81}{27} mol = 3 mol$

Given moles of oxygen $= \frac{{Given mass of O}_{2}}{{Molar mass of O}_{2}} = \frac{128}{32} mol = 4 mol$

$4 Al + 3 O_{2} ⟶ 2 {Al}_{2} O_{3}$

As per stoichiometry of the reaction, 3 mol $O_{2}$ requires 4 mol Al for complete reaction. So, 4 mol of $O_{2}$ requires $4 \times \frac{4}{3} mol = 5.33 mol Al .$ But in the reaction only 3 mol of Al is given. So Al is the limiting reagent. Amount of ${Al}_{2} O_{3}$ formed depends upon the amount of limiting reagent i.e. Al. As 4 mol Al gives 2 mol ${Al}_{2} O_{3}$ . 3 mol of Al will give 1.5 mol of ${Al}_{2} O_{3}$ . Mass of ${Al}_{2} O_{3}$ = number of moles of ${Al}_{2} O_{3}$ $\times$ molar mass of ${Al}_{2} O_{3}$ $= 1.5 \times 102 g = 153 g .$

Q 14 :

X g of benzoic acid on reaction with aq ${NaHCO}_{3}$ released ${CO}_{2}$ that occupied 11.2 L volume at STP.

X is _______ g. [2025]

(61)

Moles of $C O_{2} (n_{{CO}_{2}})$ $= \frac{Volume at STP (in L)}{22.4} = \frac{11.2}{22.4} = 0.5 mol$

$C_{6} H_{5} COOH + {NaHCO}_{3} ⟶ C_{6} H_{5} COONa + H_{2} O + {CO}_{2}$

By stoichiometry of the reaction, moles of benzoic acid $(n_{C_{6} H_{5} COOH}) = n_{{CO}_{2}} = 0.5$ mol

Molar mass of benzoic acid $(M_{C_{6} H_{5} COOH})$
$= (12 \times 7 + 16 \times 2 + 1 \times 6) {g mol}^{- 1} = 122 {g mol}^{- 1}$

Mass of benzoic acid $= n_{C_{6} H_{5} COOH} \times M_{C_{6} H_{5} COOH} = 0.5 mol \times 122 {g mol}^{- 1} = 61 g$

Q 15 :

Consider the following reaction occurring in the blast furnace.

${Fe}_{3} O_{4} (s) + 4 CO (g) \to 3 Fe (l) + 4 {CO}_{2} (g)$

$' x'$ kg of iron is produced when $2.32 \times 10^{3} {kg Fe}_{3} O_{4}$ and $2.8 \times 10^{2} kg CO$ are brought together in the furnace. The value of $' x'$ is _______ (nearest integer).

[Given:
${Molar mass of Fe}_{3} O_{4} = 232 {g mol}^{- 1}$
$Molar mass of CO = 28 {g mol}^{- 1}$
$Molar mass of Fe = 56 {g mol}^{- 1}$ ] [2025]

(420)

Moles of ${Fe}_{3} O_{4} (n_{{Fe}_{3} O_{4}})$ $= \frac{{Given mass of Fe}_{3} O_{4}}{{Molar mass of Fe}_{3} O_{4}}$

$= \frac{2.32 \times 10^{3} kg}{232 {g mol}^{- 1}} = \frac{2.32 \times 10^{3} \times 10^{3} g}{232 {g mol}^{- 1}} = 10000 mol$

Moles of CO $(n_{CO})$ $= \frac{Given mass of CO}{Molar mass of CO}$

$= \frac{2.8 \times 10^{2} kg}{28 {g mol}^{- 1}} = \frac{2.8 \times 10^{2} \times 10^{3} g}{28 {g mol}^{- 1}} = 10000 mol$

${Fe}_{3} O_{4} + 4 CO ⟶ 3 Fe + 4 {CO}_{2}$

As per stoichiometry of the reaction, 10000 mol ${Fe}_{3} O_{4}$ needs 40000 mol of CO for complete reaction. But only 10000 mol CO is present. So CO is the limiting reagent.
So amount of iron formed depends upon the amount of CO.

By reaction stoichiometry:

${Fe}_{3} O_{4} + 4 CO ⟶ 3 Fe + 4 {CO}_{2}$

$\frac{n_{Fe}}{3} = \frac{n_{CO}}{4}$

$\frac{n_{Fe}}{3} = \frac{10000}{4}$

$n_{Fe} = 3 \times \frac{10000}{4} = 7500 mol$

Mass of Fe $= {}^{n}{Fe} \times Molar mass of Fe = 7500 \times 56 g = 420000 g = 420 kg$

Q 16 :

The amount of calcium oxide produced on heating 150 kg limestone (75% pure) is _____ kg.

(Nearest integer)

Given: Molar mass (in g ${mol}^{- 1}$ ) of Ca–40, O–16, C–12 [2025]

(63)

${Mass of CaCO}_{3} in 150 kg sample of limestone (W_{{CaCO}_{3}}) = 75 % of 150 kg$ $= 0.75 \times 150 kg = 112.5 kg = 112500 g$

${Molar mass of CaCO}_{3} (M_{{CaCO}_{3}}) = 100 g/mol$

${Moles of CaCO}_{3} (n_{{CaCO}_{3}}) = \frac{W_{{CaCO}_{3}}}{M_{{CaCO}_{3}}}$

$= \frac{112500}{100} = 1125 mol$

Decomposition of calcium carbonate is represented as:

${CaCO}_{3} (s) \overset{Δ}{\to} CaO (s) + {CO}_{2} (g)$

$By stoichiometry,$

$n_{CaO} = n_{{CaCO}_{3}} = 1125 mol$

$Mass of CaO = n_{CaO} \times M_{CaO}$ $= 1125 \times 56 g = 63000 g = 63 kg$

Q 17 :

An organic compound weighing 500 mg, produced 220 mg of ${CO}_{2}$ on complete combustion. The percentage composition of carbon in the compound is ................. %. (nearest integer)

(Given molar mass in g ${mol}^{- 1}$ of C : 12, O : 16) [2025]

(12)

${Mass of CO}_{2} produced (W_{C O_{2}}) = 220 mg = 0.22 g$

${Moles of CO}_{2} produced (n_{C O_{2}}) = \frac{W_{C O_{2}}}{M_{C O_{2}}} = \frac{0.22}{44} g = 0.005 mol$

$Moles of carbon in organic compound (n_{C}) = {Moles of CO}_{2} produced (n_{C O_{2}}) = 0.005 mol$

$Mass of carbon in organic compound (W_{C}) = n_{C} \times M_{C}$ $= 0.005 \times 12 g = 0.06 g$

$Percentage of C in organic compound = \frac{W_{C}}{W_{organic compound}} \times 100$ $= \frac{0.06 g}{0.5 g} \times 100 = 12 %$

Q 18 :

Butane reacts with oxygen to produce carbon dioxide and water following the equation given below

$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 {CO}_{2} (g) + 5 H_{2} O (l)$

If 174.0 kg of butane is mixed with 320.0 kg of $O_{2}$ , the volume of water formed in litres is _____. (Nearest integer)

[Given: (a) Molar mass of C, H, O are 12, 1, 16 g ${mol}^{- 1}$ respectively, (b) Density of water = 1 g ${mL}^{- 1}$ ] [2025]

(138)

Molar mass of butane $(C_{4} H_{10})$ $= (4 \times 12 + 10 \times 1) g/mol = 58 g/mol$

$Moles of butane (n_{C_{4} H_{10}}) = \frac{Mass of butane}{Molar mass of butane}$ $= \frac{174000}{{58 g mol}^{- 1}} = 3 \times 10^{3} mol$

$Moles of oxygen (n_{O_{2}}) = \frac{Mass of oxygen}{Molar mass of oxygen}$ $= \frac{320000 g}{{32 g mol}^{- 1}} = 10 \times 10^{3} mol$

$C_{4} H_{10} + 6.5 O_{2} \to 4 {CO}_{2} + 5 H_{2} O$

As per reaction stoichiometry, to react with $3 \times 10^{3}$ mol of butane, $6.5 \times 3 \times 10^{3} mol$ i.e $19.5 \times 10^{3}$ mol oxygen is needed. But available oxygen is $10 \times 10^{3}$ mol.
Thus, oxygen is the limiting reagent. Amount of water produced depends upon amount of limiting reagent, i.e. oxygen.

As per reaction stoichiometry:

$\frac{n_{H_{2} O}}{5} = \frac{n_{O_{2}}}{6.5}$

$\frac{n_{H_{2} O}}{5} = \frac{10 \times 10^{3}}{6.5}$

$n_{H_{2} O} = \frac{5 \times 10 \times 10^{3}}{6.5} = 7692.3 mol$

$Mass of water (W_{H_{2} O}) = n_{H_{2} O} \times M_{H_{2} O}$ $= 7692.3 \times 18 g = 138461.4 g$

$Density of water (d_{H_{2} O}) = 1 g/mL$

$Volume of water = \frac{W_{H_{2} O}}{d_{H_{2} O}} = \frac{138461.4 g}{1 {g mL}^{- 1}}$ $= 138461.4 mL = 138.46 L \approx 138 L$

Q 19 :

20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value)

(Given: Na = 23, $I$ = 127, Ag = 108, N = 14, O = 16 g ${mol}^{- 1}$ ) [2025]

(1)

$Molar mass of silver iodide (M_{AgI}) = (108 + 127) g/mol = 235 g/mol$

$Moles of AgI (n_{AgI}) = \frac{W_{AgI}}{M_{AgI}} = \frac{4.74}{235} mol$

$NaI(aq) + {AgNO}_{3} (aq)(excess) \to AgI(s) + {NaNO}_{3} (aq)$

As per reaction stoichiometry:

$Moles of NaI (n_{NaI}) = Moles of AgI (n_{AgI}) = \frac{4.74}{235} mol$

$Volume of NaI solution (V_{NaI}) = 20 mL = 0.02 L$

$Molarity of NaI (M_{NaI}) = \frac{n_{NaI}}{V_{NaI} (in L)} = \frac{4.74}{235 \times 0.02} mol$

$= 1.008 M$

Q 20 :

X g of nitrobenzene on nitration gave 4.2 g of m-dinitrobenzene.

X = ______ g. (nearest integer)

[Given: molar mass (in g ${mol}^{- 1}$ ) C : 12, H : 1, O : 16, N : 14] [2025]

(3)

$Moles of m -dinitrobenzene = \frac{Mass of m -dinitrobenzene}{Molar mass of m -dinitrobenzene}$

$= \frac{4.2}{168} mol = 0.025 mol$

$As per reaction stoichiometry, moles of nitrobenzene = moles of m -dinitrobenzene = 0.025 mol$

$Mass of nitrobenzene = moles of nitrobenzene \times molar mass of nitrobenzene$

$= 0.025 \times 123 g = 3.075 g$

Q 21 :

Some ${CO}_{2}$ gas was kept in a sealed container at a pressure of 1 atm and at 273 K. This entire amount of ${CO}_{2}$ gas was later passed through an aqueous solution of $Ca {(OH)}_{2}$ . The excess unreacted $Ca {(OH)}_{2}$ was later neutralised with 0.1 M of 40 mL HCl. If the volume of the sealed container of ${CO}_{2}$ was $x$ , then $x$ is ______ ${cm}^{3}$ (nearest integer).

[Given: The entire amount of ${CO}_{2} (g)$ reacted with exactly half the initial amount of $Ca {(OH)}_{2}$ present in the aqueous solution.] [2025]

(45)

$Moles of HCl consumed = molarity of HCl \times volume of HCl$

$= 0.1 M \times 40 mL = 4 mmol$

${Ca(OH)}_{2} + 2 HCl \to {CaCl}_{2} + 2 H_{2} O$

${By stoichiometry, moles of Ca(OH)}_{2} that reacted with HCl = \frac{Moles of HCl}{2} = \frac{4 mmol}{2} = 2 mmol$

As out of total $Ca {(OH)}_{2}$ present initially, half reacted with HCl and the other half reacted with ${CO}_{2}$ .

${Moles of Ca(OH)}_{2} {reacted with CO}_{2} = {Moles of Ca(OH)}_{2} reacted with HCl = 2 mmol$

${CO}_{2} + {Ca(OH)}_{2} ⟶ {CaCO}_{3} + H_{2} O$

${By stoichiometry, moles of CO}_{2} = {moles of Ca(OH)}_{2} {reacted with CO}_{2} = 2 mmol = 0.002 mol$

At 1 atm and 273 K, 1 mol of any gas occupies $22400$ ${cm}^{3}$ . So $0.002 {mol of CO}_{2} occupy 0.002 \times 22400$ ${cm}^{3}$

$= 44.8 mL \approx 45 mL$

Q 22 :

The volume of 0.02 M aqueous HBr required to neutralize 10.0 mL of 0.01 M aqueous $Ba {(OH)}_{2}$ is (Assume complete neutralization) [2023]

2.5 mL
7.5 mL
5.0 mL
10.0 mL

1

2

3

4

(4)

${Ba(OH)}_{2} + 2 HBr \to {BaBr}_{2} + 2 H_{2} O$

$mmol$ $0.1$

$Required mmol of HBr = 0.2 = 0.02 \times V_{mL}$

$V_{mL} = 10 mL = 1.362 L = 1362 mL$

Q 23 :

5 g of NaOH was dissolved in deionized water to prepare a 450 mL stock solution. What volume (in mL) of this solution would be required to prepare 500 mL of 0.1 M solution? _________. Given: Molar Mass of Na, O and H is 23, 16 and 1 g ${mol}^{- 1}$ respectively. [2023]

(180)

${(Molarity)}_{i} = {\frac{5 \times 1000}{40 \times 450}}$

$By dilution law M_{i} V_{i} = M_{f} V_{f}$

$\Rightarrow {\frac{5 \times 1000}{40 \times 450}} \times V = 0.1 \times 500$

$V = \frac{50 \times 40 \times 450}{5 \times 1000} = 5 \times 4 \times 9 = 180 mL$

Q 24 :

When 0.01 mol of an organic compound containing 60% carbon was burnt completely, 4.4 g of ${CO}_{2}$ was produced. The molar mass of compound is ______ g ${mol}^{- 1}$ (Nearest integer). [2023]

(200)

$Let M = Molar mass of the compound$

$Mass of carbon = 0.01 M \times \frac{60}{100}$

$Mole of carbon = \frac{0.01 M}{12} \times \frac{60}{100}$

${Mole of CO}_{2} from combustion = \frac{4.4}{44} = mole of carbon$

$\frac{0.01 M}{12} \times \frac{60}{100} = \frac{4.4}{44}$ $\Rightarrow M = \frac{4.4}{44} \times \frac{100}{60} \times \frac{12}{0.01} = 200 g/mol$

Q 25 :

Some amount of dichloromethane $({CH}_{2} {Cl}_{2})$ is added to 671.141 mL of chloroform $({CHCl}_{3})$ to prepare $2.6 \times 10^{- 3}$ M solution of ${CH}_{2} {Cl}_{2}$ (DCM). The concentration of DCM is ______ ppm (by mass). Given: atomic mass: C = 12, H = 1, Cl = 35.5; density of ${CHCl}_{3}$ = 1.49 g ${cm}^{- 3}$ . [2023]

(148)

$Molarity = \frac{W \times 1000}{M_{w} \times V}$

$2.6 \times 10^{- 3} = \frac{W \times 1000}{85 \times 671.141}$

$W = \frac{2.6 \times 10^{- 3} \times 85 \times 671.141}{1000} = 0.148 g$

${Conc}^{n} . of DCM (in ppm) = \frac{0.148}{1.49 \times 671.141} \times 10^{6} = 148 ppm$

Q 26 :

On complete combustion, 0.492 g of an organic compound gave 0.792 g of $C O_{2}$ . The % of carbon in the organic compound is _______. (Nearest Integer) [2023]

(44)

$44 gm of {CO}_{2} contains 12 g carbon$

$0.792 gm of {CO}_{2} contains \frac{0.792 \times 12}{44} g of carbon$

$% of carbon = \frac{0.216}{0.492} \times 100 = 43.9 % \approx 44 %$

Q 27 :

Assume carbon burns according to following equation:

$2 C_{(s)} + O_{2 (g)} \to 2 {CO}_{(g)}$

When 12 g carbon is burnt in 48 g of oxygen, the volume of carbon monoxide produced is ______ $\times 10^{- 1}$ L at STP (Nearest Integer).

[Given: Assume CO as ideal gas, Mass of C is 12 g ${mol}^{- 1}$ , Mass of O is 16 g ${mol}^{- 1}$ and molar volume of an ideal gas at STP is 22.7 L ${mol}^{- 1}$ ] [2023]

(227)

$2 C$ $+$ $O_{2}$ $⟶$ $2 C O$

$\frac{12}{12} (LR is C)$ $\frac{48}{32}$

$1 mole$ $1.5 mole$

$0$ $\begin{matrix} (1.5 - 0.5) \\ = 1.0 mole \end{matrix}$ $1 mole$

$Mole of CO = 1 mole$

${(V_{C O})}_{at STP} = 22.7 Lit = 227 \times 10^{- 1} Lit$

Q 28 :

The molality of a 10% $(v / v)$ solution of di-bromine solution in ${CCl}_{4}$ (carbon tetrachloride) is $‘ x ’$ .

$x =______\times 10^{- 2} m$ . (Nearest integer)

[Given:
molar mass of ${Br}_{2}$ = 160 g ${mol}^{- 1}$
atomic mass of C = 12 g ${mol}^{- 1}$
atomic mass of Cl = 35.5 g ${mol}^{- 1}$
density of dibromine = 3.2 g ${cm}^{- 3}$
density of ${CCl}_{4}$ = 1.6 g ${cm}^{- 3}$ ] [2023]

(139)

$Molarity 10 % = \frac{V}{V}$

Solvent = ${CCl}_{4}$ , Solute = ${Br}_{2}$

100 mL has 10 mL ${Br}_{2}$ and 90 mL ${CCl}_{4}$

$m = 10 \times 3.0 = 32 g$

$n = \frac{32}{160} = \frac{1}{5} mol$

$m_{{CCl}_{4}} = 90 \times 1.6$

$= 144 g = 0.144 kg$

$m = \frac{1 / 5}{0.144} = 1.39 = 139 \times 10^{- 2}$

Q 29 :

If 5 moles of ${BaCl}_{2}$ is mixed with 2 moles of ${Na}_{3} {PO}_{4}$ , the maximum number of moles of ${Ba}_{3} {({PO}_{4})}_{2}$ formed is ______ (Nearest integer) [2023]

(1)

$\begin{matrix} 3 {BaCl}_{2} & + & 2 {Na}_{3} {PO}_{4} & \to & {Ba}_{3} {({PO}_{4})}_{2} & + & 6 NaCl \\ Initial moles & 5 & 2 \end{matrix}$

$L.R.$ $\frac{5}{3} = 1.67$ $\frac{2}{2} = 1 (L.R. = {Na}_{3} {PO}_{4})$

$\frac{{Moles of Na}_{3} {PO}_{4}}{2} = \frac{{Moles of Ba}_{3} {({PO}_{4})}_{2}}{1}$

${Maximum number of moles of Ba}_{3} {({PO}_{4})}_{2} = \frac{2}{2} = 1 mole$

Q 30 :

The volume of hydrogen liberated at STP by treating 2.4 g magnesium with excess of hydrochloric acid is ______ $\times 10^{- 2}$ L [2023]

Given: Molar volume of gas is 22.4 L at STP.

Molar mass of magnesium is 24 g ${mol}^{- 1}$

(224)

$Mg + 2 HCl \to {MgCl}_{2} + H_{2} ↑$

$1 mole 1 mole$

$w = 2.4 g$

$= 0.1 mole 0.1 mole$

$1 mole of gas at STP$

$= 22.4 lit.$

$∴$ $0.1 mole of gas = 0.1 \times 22.4$

$= 2.24 lit = 224 \times 10^{- 2} litre$

Topic Question Set

Q 1 :

A sample of $C a C O_{3}$ and $M g C O_{3}$ weighed 2.21 g is ignited to constant weight of 1.152 g. The composition of mixture is

(Given molar mass in g $m o l^{- 1}$ $C a C O_{3} : 100, M g C O_{3} : 84$ ) [2024]

Q 2 :

The number of moles of methane required to produce 11g $C O_{2}$ (g) after complete combustion is:

[Given molar mass of methane in g $m o l^{- 1}$ : 16] [2024]

Q 3 :

Combustion of glucose $(C_{6} H_{12} O_{6})$ produces $C O_{2}$ and water. The amount of oxygen (in g) required for the complete combustion of 900g of glucose is: [Molar mass of glucose in g $m o l^{- 1}$ = 180] [2024]

Q 4 :

Mass of methane required to produce 22 g of $C O_{2}$ after complete combustion is __________ g.

(Given Molar mass in g $m o l^{- 1}$ ,C = 12.0, H = 1.0, O = 16.0) [2024]

Q 5 :

If 50 mL of 0.5 M oxalic acid is required to neutralize 25 mL of NaOH solution, the amount of NaOH in 50 mL of given NaOH solution is _________ g. [2024]

Q 6 :

Number of moles of methane required to produce 22g $C O_{2 (g)}$ after combustion is $x \times 10^{- 2}$ moles. The value of $x$ is _________ . [2024]

Q 8 :

10 mL of gaseous hydrocarbon on combustion gives 40 mL of $C O_{2} (g)$ and 50 mL of water vapour. Total number of carbon and hydrogen atoms in the hydrocarbon is ________ [2024]

Q 9 :

9.3 g of aniline is subjected to reaction with excess of acetic anhydride to prepare acetanilide. The mass of acetanilide produced if the reaction is 100% completed is ______ $\times 10^{- 1}$ g.

(Given molar mass in g ${mol}^{- 1}$ N : 14, O : 16
C : 12, H : 1) [2024]

Q 12 :

Mass of magnesium required to produce 220 mL of hydrogen gas at STP on reaction with excess of dil. HCl is

Given: Molar mass of Mg is 24 g ${mol}^{- 1}$ . [2025]

Q 13 :

When 81.0 g of aluminium is allowed to react with 128.0 g of oxygen gas, the mass of aluminium oxide produced in grams is _______. (Nearest integer)

Given:
Molar mass of Al is 27.0 g ${mol}^{- 1}$
Molar mass of O is 16.0 g ${mol}^{- 1}$ [2025]

Q 14 :

X g of benzoic acid on reaction with aq ${NaHCO}_{3}$ released ${CO}_{2}$ that occupied 11.2 L volume at STP.

X is _______ g. [2025]

Q 16 :

The amount of calcium oxide produced on heating 150 kg limestone (75% pure) is _____ kg.

(Nearest integer)

Given: Molar mass (in g ${mol}^{- 1}$ ) of Ca–40, O–16, C–12 [2025]

Q 17 :

An organic compound weighing 500 mg, produced 220 mg of ${CO}_{2}$ on complete combustion. The percentage composition of carbon in the compound is ................. %. (nearest integer)

(Given molar mass in g ${mol}^{- 1}$ of C : 12, O : 16) [2025]

Q 19 :

20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value)

(Given: Na = 23, $I$ = 127, Ag = 108, N = 14, O = 16 g ${mol}^{- 1}$ ) [2025]

Q 20 :

X g of nitrobenzene on nitration gave 4.2 g of m-dinitrobenzene.

X = ______ g. (nearest integer)

[Given: molar mass (in g ${mol}^{- 1}$ ) C : 12, H : 1, O : 16, N : 14] [2025]

Q 22 :

The volume of 0.02 M aqueous HBr required to neutralize 10.0 mL of 0.01 M aqueous $Ba {(OH)}_{2}$ is (Assume complete neutralization) [2023]

Q 23 :

5 g of NaOH was dissolved in deionized water to prepare a 450 mL stock solution. What volume (in mL) of this solution would be required to prepare 500 mL of 0.1 M solution? _________. Given: Molar Mass of Na, O and H is 23, 16 and 1 g ${mol}^{- 1}$ respectively. [2023]

Q 24 :

When 0.01 mol of an organic compound containing 60% carbon was burnt completely, 4.4 g of ${CO}_{2}$ was produced. The molar mass of compound is ______ g ${mol}^{- 1}$ (Nearest integer). [2023]

Q 26 :

On complete combustion, 0.492 g of an organic compound gave 0.792 g of $C O_{2}$ . The % of carbon in the organic compound is _______. (Nearest Integer) [2023]

Q 29 :

If 5 moles of ${BaCl}_{2}$ is mixed with 2 moles of ${Na}_{3} {PO}_{4}$ , the maximum number of moles of ${Ba}_{3} {({PO}_{4})}_{2}$ formed is ______ (Nearest integer) [2023]

Q 30 :

The volume of hydrogen liberated at STP by treating 2.4 g magnesium with excess of hydrochloric acid is ______ $\times 10^{- 2}$ L [2023]

Given: Molar volume of gas is 22.4 L at STP.

Molar mass of magnesium is 24 g ${mol}^{- 1}$

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Topic Question Set

Q 1 : A sample of CaCO3 and MgCO3 weighed 2.21 g is ignited to constant weight of 1.152 g. The composition of mixture is (Given molar mass in g mol-1 CaCO3:100,MgCO3:84) [2024]

Q 2 : The number of moles of methane required to produce 11g CO2(g) after complete combustion is: [Given molar mass of methane in g mol-1 : 16] [2024]

Q 3 : Combustion of glucose (C6H12O6) produces CO2 and water. The amount of oxygen (in g) required for the complete combustion of 900g of glucose is: [Molar mass of glucose in g mol-1= 180] [2024]

Q 4 : Mass of methane required to produce 22 g of CO2 after complete combustion is __________ g. (Given Molar mass in g mol-1,C = 12.0, H = 1.0, O = 16.0) [2024]

Q 5 : If 50 mL of 0.5 M oxalic acid is required to neutralize 25 mL of NaOH solution, the amount of NaOH in 50 mL of given NaOH solution is _________ g. [2024]

Q 6 : Number of moles of methane required to produce 22g CO2(g) after combustion is x×10-2 moles. The value of x is _________ . [2024]

Q 7 : Consider the following reaction: 3PbCl2+2(NH4)3PO4→Pb3(PO4)2+6NH4Cl If 72 mmol of PbCl2 is mixed with 50 mmol of (NH4)3PO4, then the amount of Pb3(PO4)2 formed is _______ mmol (nearest integer). [2024]

Q 8 : 10 mL of gaseous hydrocarbon on combustion gives 40 mL of CO2(g) and 50 mL of water vapour. Total number of carbon and hydrogen atoms in the hydrocarbon is ________ [2024]

Q 9 : 9.3 g of aniline is subjected to reaction with excess of acetic anhydride to prepare acetanilide. The mass of acetanilide produced if the reaction is 100% completed is ______ ×10-1 g. (Given molar mass in g mol-1 N : 14, O : 16 C : 12, H : 1) [2024]

Q 10 : CaCO3(s)+2HCl(aq) → CaCl2(aq)+CO2(g)+H2O(l) Consider the above reaction, what mass of CaCl2 will be formed if 250 mL of 0.76 M HCl reacts with 1000 g of CaCO3 ? (Given: Molar mass of Ca, C, O, H and Cl are 40, 12, 16, 1 and 35.5 g mol-1, respectively) [2025]

Q 12 : Mass of magnesium required to produce 220 mL of hydrogen gas at STP on reaction with excess of dil. HCl is Given: Molar mass of Mg is 24 g mol-1. [2025]

Q 13 : When 81.0 g of aluminium is allowed to react with 128.0 g of oxygen gas, the mass of aluminium oxide produced in grams is _______. (Nearest integer) Given: Molar mass of Al is 27.0 g mol-1 Molar mass of O is 16.0 g mol-1 [2025]

Q 14 : X g of benzoic acid on reaction with aq NaHCO3 released CO2 that occupied 11.2 L volume at STP. X is _______ g. [2025]

Q 16 : The amount of calcium oxide produced on heating 150 kg limestone (75% pure) is _____ kg. (Nearest integer) Given: Molar mass (in g mol-1) of Ca–40, O–16, C–12 [2025]

Q 17 : An organic compound weighing 500 mg, produced 220 mg of CO2 on complete combustion. The percentage composition of carbon in the compound is ................. %. (nearest integer) (Given molar mass in g mol-1 of C : 12, O : 16) [2025]

Q 19 : 20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value) (Given: Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol-1) [2025]

Q 20 : X g of nitrobenzene on nitration gave 4.2 g of m-dinitrobenzene. X = ______ g. (nearest integer) [Given: molar mass (in g mol-1) C : 12, H : 1, O : 16, N : 14] [2025]

Q 22 : The volume of 0.02 M aqueous HBr required to neutralize 10.0 mL of 0.01 M aqueous Ba(OH)2 is (Assume complete neutralization) [2023]

Q 23 : 5 g of NaOH was dissolved in deionized water to prepare a 450 mL stock solution. What volume (in mL) of this solution would be required to prepare 500 mL of 0.1 M solution? _________. Given: Molar Mass of Na, O and H is 23, 16 and 1 g mol-1 respectively. [2023]

Q 24 : When 0.01 mol of an organic compound containing 60% carbon was burnt completely, 4.4 g of CO2 was produced. The molar mass of compound is ______ g mol-1 (Nearest integer). [2023]

Q 25 : Some amount of dichloromethane (CH2Cl2) is added to 671.141 mL of chloroform (CHCl3) to prepare 2.6×10-3 M solution of CH2Cl2 (DCM). The concentration of DCM is ______ ppm (by mass). Given: atomic mass: C = 12, H = 1, Cl = 35.5; density of CHCl3 = 1.49 g cm-3. [2023]

Q 26 : On complete combustion, 0.492 g of an organic compound gave 0.792 g of CO2. The % of carbon in the organic compound is _______. (Nearest Integer) [2023]

Q 29 : If 5 moles of BaCl2 is mixed with 2 moles of Na3PO4, the maximum number of moles of Ba3(PO4)2 formed is ______ (Nearest integer) [2023]

Q 30 : The volume of hydrogen liberated at STP by treating 2.4 g magnesium with excess of hydrochloric acid is ______ ×10-2 L [2023] Given: Molar volume of gas is 22.4 L at STP. Molar mass of magnesium is 24 g mol-1

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Q 1 :

A sample of $C a C O_{3}$ and $M g C O_{3}$ weighed 2.21 g is ignited to constant weight of 1.152 g. The composition of mixture is

(Given molar mass in g $m o l^{- 1}$ $C a C O_{3} : 100, M g C O_{3} : 84$ ) [2024]

Q 2 :

The number of moles of methane required to produce 11g $C O_{2}$ (g) after complete combustion is:

[Given molar mass of methane in g $m o l^{- 1}$ : 16] [2024]

Q 3 :

Combustion of glucose $(C_{6} H_{12} O_{6})$ produces $C O_{2}$ and water. The amount of oxygen (in g) required for the complete combustion of 900g of glucose is: [Molar mass of glucose in g $m o l^{- 1}$ = 180] [2024]

Q 4 :

Mass of methane required to produce 22 g of $C O_{2}$ after complete combustion is __________ g.

(Given Molar mass in g $m o l^{- 1}$ ,C = 12.0, H = 1.0, O = 16.0) [2024]

Q 5 :

If 50 mL of 0.5 M oxalic acid is required to neutralize 25 mL of NaOH solution, the amount of NaOH in 50 mL of given NaOH solution is _________ g. [2024]

Q 6 :

Number of moles of methane required to produce 22g $C O_{2 (g)}$ after combustion is $x \times 10^{- 2}$ moles. The value of $x$ is _________ . [2024]

Q 8 :

10 mL of gaseous hydrocarbon on combustion gives 40 mL of $C O_{2} (g)$ and 50 mL of water vapour. Total number of carbon and hydrogen atoms in the hydrocarbon is ________ [2024]

Q 9 :

9.3 g of aniline is subjected to reaction with excess of acetic anhydride to prepare acetanilide. The mass of acetanilide produced if the reaction is 100% completed is ______ $\times 10^{- 1}$ g.

(Given molar mass in g ${mol}^{- 1}$ N : 14, O : 16
C : 12, H : 1) [2024]

Q 12 :

Mass of magnesium required to produce 220 mL of hydrogen gas at STP on reaction with excess of dil. HCl is

Given: Molar mass of Mg is 24 g ${mol}^{- 1}$ . [2025]

Q 13 :

When 81.0 g of aluminium is allowed to react with 128.0 g of oxygen gas, the mass of aluminium oxide produced in grams is _______. (Nearest integer)

Given:
Molar mass of Al is 27.0 g ${mol}^{- 1}$
Molar mass of O is 16.0 g ${mol}^{- 1}$ [2025]

Q 14 :

X g of benzoic acid on reaction with aq ${NaHCO}_{3}$ released ${CO}_{2}$ that occupied 11.2 L volume at STP.

X is _______ g. [2025]

Q 16 :

The amount of calcium oxide produced on heating 150 kg limestone (75% pure) is _____ kg.

(Nearest integer)

Given: Molar mass (in g ${mol}^{- 1}$ ) of Ca–40, O–16, C–12 [2025]

Q 17 :

An organic compound weighing 500 mg, produced 220 mg of ${CO}_{2}$ on complete combustion. The percentage composition of carbon in the compound is ................. %. (nearest integer)

(Given molar mass in g ${mol}^{- 1}$ of C : 12, O : 16) [2025]

Q 19 :

20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value)

(Given: Na = 23, $I$ = 127, Ag = 108, N = 14, O = 16 g ${mol}^{- 1}$ ) [2025]

Q 20 :

X g of nitrobenzene on nitration gave 4.2 g of m-dinitrobenzene.

X = ______ g. (nearest integer)

[Given: molar mass (in g ${mol}^{- 1}$ ) C : 12, H : 1, O : 16, N : 14] [2025]

Q 22 :

The volume of 0.02 M aqueous HBr required to neutralize 10.0 mL of 0.01 M aqueous $Ba {(OH)}_{2}$ is (Assume complete neutralization) [2023]

Q 23 :

5 g of NaOH was dissolved in deionized water to prepare a 450 mL stock solution. What volume (in mL) of this solution would be required to prepare 500 mL of 0.1 M solution? _________. Given: Molar Mass of Na, O and H is 23, 16 and 1 g ${mol}^{- 1}$ respectively. [2023]

Q 24 :

When 0.01 mol of an organic compound containing 60% carbon was burnt completely, 4.4 g of ${CO}_{2}$ was produced. The molar mass of compound is ______ g ${mol}^{- 1}$ (Nearest integer). [2023]

Q 26 :

On complete combustion, 0.492 g of an organic compound gave 0.792 g of $C O_{2}$ . The % of carbon in the organic compound is _______. (Nearest Integer) [2023]

Q 29 :

If 5 moles of ${BaCl}_{2}$ is mixed with 2 moles of ${Na}_{3} {PO}_{4}$ , the maximum number of moles of ${Ba}_{3} {({PO}_{4})}_{2}$ formed is ______ (Nearest integer) [2023]

Q 30 :

The volume of hydrogen liberated at STP by treating 2.4 g magnesium with excess of hydrochloric acid is ______ $\times 10^{- 2}$ L [2023]

Given: Molar volume of gas is 22.4 L at STP.

Molar mass of magnesium is 24 g ${mol}^{- 1}$