Question

If the standard molar enthalpy change for combustion of graphite powder is $-2.48\times {10}^{2}kJ{\text{mol}}^{-1}.$ the amount of heat generated on combustion of $1g$ of graphite powder in $kJ$ is
(Nearest integer)

Answer 1

$\text{Heat of combustion per mol for}12g=-2.48\times {10}^{2}KJ$
For $1g$ of graphite :
$\text{Heat of combustion}=\frac{-2.48\times {10}^2}{12}=-20.66kJ\approx -21$