A cylinder of gas is assumed to contain 11.2 | vedclass

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(C) Molar mass of butane $(C_4H_{10})$ $= (4 	imes 12) + (10 	imes 1) = 58 \ g/mol$. Number of moles of butane in $11.2 \ kg$ $(11200 \ g)$ $= \frac

Vedclass · Accepted Answer

$26$

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(C) Molar mass of butane $(C_4H_{10})$ $= (4 	imes 12) + (10 	imes 1) = 58 \ g/mol$. Number of moles of butane in $11.2 \ kg$ $(11200 \ g)$ $= \frac{11200 \ g}{58 \ g/mol} \approx 193.1 \ mol$. Total energy released by combustion of $11.2 \ kg$ of butane $= 193.1 \ mol 	imes 2658 \ kJ/mol \approx 513260 \ kJ$. Daily energy requirement $= 20000 \ kJ$. Number of days the cylinder will last $= \frac{513260 \ kJ}{20000 \ kJ/day} \approx 25.66 \ days$. Rounding to the nearest whole number,the cylinder will last for $26 \ days$.

$A$ cylinder of gas is assumed to contain $11.2 \ kg$ of butane $(C_4H_{10})$. If a normal family needs $20000 \ kJ$ of energy per day,the cylinder will last: (Given that $\Delta H$ for combustion of butane is $-2658 \ kJ/mol$)....$days$

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