Question
The heat of combustion of ethane gas is $373$ kcal. Per mole. Assuming that $50\%$ of heat is lost, how many litres of ethane measured at STP must be burnt to convert $50g$ of water at $10°C$ to steam at $100°C$? One mole of gas occupies $22.4$ litres at STP. Take latent heat of steam = $2.25 \times 10^6J/g^{-1}$.

Answer

Total heat energy required to convert 50g of water at $10^\circ C$ to steam at $100^\circ C$, $=\text{cm}\Delta\text{T}+\text{mL}$
$=1000\times50\times(100-10)+\frac{50\times2.25\times10^6}{4.2}$
$=4.5\times10^6+26.79\times10^6$
$=31.29\times10^6\text{ cal}$ As 50% of heat is lost, $\therefore$ Total heat product = $2 \times 31.29 \times 10^6cal$ Heat of combustion = $373 \times 10^3cal/mole$
$\therefore$ No. of mole of ethane to be burnt, $=\frac{2\times31.29\times10^6}{373\times10^3}\text{mole}$ Volume of ethane, $=\frac{2\times31.29\times10^6}{373\times10^3}$
$=3758.2\text{ liters}$

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