Question

The heat of combustion of sucrose $\left(C_{12}{H}_{22}{O}_{11}\right)$ is $-5.65\times {10}^{3}kJ$. If complete combustion of $1.0kg$ sucrose is done, heat evloved will be:

• A. $1.65\times {10}^{4}kJ$
• B. $5.65\times {10}^{3}kJ$
• C. $7.38\times {10}^{5}kJ$
• D. $3.51\times {10}^{5}kJ$

Answer 1

The correct option is A $1.65\times {10}^{4}kJ$

Molecular weight of $C_{12}{H}_{22}{O}_{11}$
$=(12\times 12+22+11\times 16)g$
$=(144+22+176)g=342g$
$\because \text{Heat evolved on combustion of}342g\text{sucrose}=5.65\times {10}^{3}kJ$
$\therefore$ Heat evolved on combustion of $1000g$ sucrose.
$=\frac{5.65\times {10}^3}{342}\times 1000=1.652\times {10}^{4}kJ$