Butane reacts with oxygen to produce carbon dioxide and water following the equation given below If 174.0 kg of butane is mixed with 320.0 kg of , the volume of water formed in litres is _____. (Nearest integer) [Given: (a) Molar mass of C, H, O are 1

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Solution

Q.
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]

Ans.
(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:

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

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

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

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

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

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

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