Question

A cylinder of gas contains 14.5 kg of butane. If a normal family requires $20,000kJ$ of energy per day for cooking, the butane gas in the cylinder lasts for how many days?
$(\Delta H_{comb}$ of $C_4{H}_{10}=-2568kJmo{l}^{-1})$

• A. 61.1
• B. 32.1
• C. 50
• D. 40

Answer 1

The correct option is B 32.1

Given that, 1 mol of (58 g) butane produces $2568kJ$ energy.
So, 14500 g of butane produces $=\frac{2568}{58}\times 14500kJ$ of energy.
Energy required per day = 20,000 kJ
Number of days $=\frac{2568}{58}\times \frac{14500}{20,000}$
= 32.1 days