Question

A cylinder of cooking gas is assumed to contain $11.2kg$ of Butane. The thermochemical equation for the combustion of butane is $C_4{H}_{10\left(s\right)}+\frac{13}{2}{O}_2\left(g\right)\to 4C{O}_{2\left(g\right)}=5h_{2}C\left(l\right)△H=-265.8KJ$.If a family needs $1500KJ$ of energy per day for cooking how long does the cylinder last.Assuming that $30\%$ of it is wasted due to incopmplete combustion. How long would the cylinder last in this case?

Answer 1

Molar mass of butane $=58g/mol$

$58g$ of butane gives $265.8kJ$ of heat energy.

$\therefore 11.2kg$ of butane will give heat energy $=\frac{265.8}{58}\times 11200=51.327\times {10}^{3}KJ$

As given that the $30\%$ of it is wasted, i.e., only $70\%$ of it is used.

Therefore,

Amount of heat enrgy used $=\frac{70}{100}\times 51.327\times {10}^3=3.6\times {10}^4$

Daily energy required for cooking $=1500\text{kJ per day}$

$\therefore$ Number of days cylinder will last $=\frac{3.6\times {10}^4}{1500}\approx 24\text{days}$

Hence the cylinder will last for $24$ days.