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(D) The molar mass of butane $(C_4H_{10})$ is $4 	imes 12 + 10 	imes 1 = 58 \, g \, mol^{-1}$.Combustion of $58 \, g$ of butane releases $2658 \, kJ

Vedclass · Accepted Answer

$34$

Vedclass · Answer

(D) The molar mass of butane $(C_4H_{10})$ is $4 	imes 12 + 10 	imes 1 = 58 \, g \, mol^{-1}$.Combustion of $58 \, g$ of butane releases $2658 \, kJ$ of heat.Total heat produced by $11.2 \, kg$ $(11200 \, g)$ of butane is:$Q = \frac{2658 \, kJ}{58 \, g} 	imes 11200 \, g = 512965.5 \, kJ$.Daily energy requirement is $15000 \, kJ$.Number of days the cylinder will last = $\frac{512965.5 \, kJ}{15000 \, kJ/day} \approx 34.2 \, days$.Thus,the cylinder will last for approximately $34$ days.

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