$29 g$ of butane is burnt in excess of $O_{2}$ at STP. Find the volume of the $C O_{2}$ gas produced at STP.

22.4 L

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44.8 L

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33.6 L

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11.2 L

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Solution

The correct option is B $44.8 L$
The balanced chemical equation is:
$C_{4} H_{10} + \frac{13}{8} O_{2} \to 4 C O_{2} + 5 H_{2} O$
No. of moles of $C_{4} H_{10} = \frac{29}{58} = 0.5 mol$
Since 1 mol of $C_{4} H_{10}$ gives 4 mol of $C O_{2}$ .
Hence 0.5 mol of $C_{4} H_{10}$ will give $\frac{4}{1} \times 0.5 = 2 mol$ of $C O_{2}$ .

Volume of $C O_{2}$ produced at STP $= 22.4 L \times 2 = 44.8 L$

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Stoichiometric Calculations

Standard XII Chemistry

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29 g of butane is burnt in excess of O2 at STP. Find the volume of the CO2 gas produced at STP.

The correct option is B 44.8 LThe balanced chemical equation is: C4H10+138O2→4CO2+5H2O No. of moles of C4H10=2958=0.5 mol Since 1 mol of C4H10 gives 4 mol of CO2. Hence 0.5 mol of C4H10 will give 41×0.5=2 mol of CO2. Volume of CO2 produced at STP =22.4 L×2=44.8 L

$29 g$ of butane is burnt in excess of $O_{2}$ at STP. Find the volume of the $C O_{2}$ gas produced at STP.