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

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

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

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

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

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

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

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

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]

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(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]

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(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]

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(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]

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(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]

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

By stoichiometry:

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

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

$\Rightarrow x = 4, y = 10$

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

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]

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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]

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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]

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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]

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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]

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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]

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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]

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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]

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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]

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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]