Question

A cylinder of gas assumed to contain $11.2kg$ of butane $C_4{H}_{10}$. If a normal family needs $20000kJ$ of energy per day. The cylinder will last: (Given that $\Delta H$ for combustion of butane is -2658 kJ)

• A. $20$ days
• B. $25$ days
• C. $26$ days
• D. $24$ days

Answer 1

Answer: (C)

$26$ days

$C_4{H}_{10}+\frac{13}{2}{O}_2⟶4C{O}_2+5H_{2}O$

Moles of $C_4{H}_{10}=\frac{11.2}{58}=0.193$ $kg-moles=193$ $g-mole$

Total Energy released $=193\times 2658$ $kJ$

No. of days $=\frac{193\times 2658}{20000}$ $\left(\frac{Totalenergy}{Energyrequiredperday}\right)=26$ days