Question

An LPG cylinder is assumed to contain $11.2$ kg of butane $\left(C_4{H}_{10}\right)$. If a family needs $20000$ kJ of energy per day, then cylinder will last for nearly how many days?

(Given that $\Delta H$ for combustion of butane is $-2658$ kJ/mol)

• A. $26$ days
• B. $25$ days
• C. $28$ days
• D. $24$ days

Answer 1

The correct option is A $26$ days

Molar mass of butane,

Implies that

$C_4{H_1}_0=12\times 4+10$

Implies that

$=58gmo{l}^{-1}$

Since,

$58g$ of butane gives $2658kJ$ of heat energy.

Given:

$11.2kg$ of butane will gives heat energy

$=\frac{2658kJ\times \left(11.2\times {10}^{3}g\right)}{58g}$

Implies that

$=513.268\times {10}^{3}kJ$

Daily energy requirement for cooking

$=20000kJ$

Implies that

$=2\times {10}^{4}kJda{y}^{-1}$

Hence,

The number of days LPG cylinder will last

$=\frac{513.268\times {10}^{3}kJ}{2\times {10}^{4}kJda{y}^{-1}}$

Implies that

$=26days$.