Question

What is the amount of hear released when $3.4$ gm $N{H}_3$(g) and $6.4$gm $O_2$(g) react at constant temperature and pressure by the following equation.
$4N{H}_3$(g) + $5O_2$ (g) \rightarrow $4NO\left(g\right)$ + $6H_{2}O$(g) . ${\Delta }_1{H}^0=900k.$

• A. $45$ kJ
• B. $250$ kJ
• C. $36$ kJ
• D. None of these

Answer 1

Answer: (C)

$36$ kJ

The molecular weights of $N{H}_3$ and $O_2$ are 17 g/mol and 32 g/mol respectively.
3.4 g $N{H}_3=\frac{3.4g}{17g/mol}=0.2mol$
6.4 g $O_2=\frac{6.4g}{32g/mol}=0.2mol$
0.2 moles of $O_2$ will react with $0.2\times \frac{4}{5}=0.16$ moles of $N{H}_3$. But 0.2 moles of
$N{H}_3$ are present. Hence, $O_2$ is the limiting reagent and $N{H}_3$ is excess reagent.

When 5 moles of $O_2$ react, the heat released is 900 kJ.

When 0.2 moles of $O_2$ react, the heat released will be

$900kJ\times \frac{0.2}{5}=36kJ$.