Question

How much heat is absorbed during production of 147 g of NO by the combination of nitrogen and oxygen?

Answer 1

Answer:

$\large \boxed{\text{105 kcal}}$

Explanation:

MM:                        30.01

            N₂ + O₂ ⟶ 2NO; ΔH = +43 kcal/mol

m/g:                         147

Treat the heat as if it were a reactant in the reaction. Then you can write

N₂ + O₂ + 43 kcal ⟶ 2NO

The conversion factor is then 43 kcal/2 mol NO.

1. Moles of NO

$\text{Moles of NO} = \text{147 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{4.898 mol NO}$

2. Amount of heat

$\text{Heat} = \text{4.898 mol NO } \times \dfrac{\text{43 kcal}}{\text{2 mol NO}} = \text{105 kcal}\\\\\text{The reaction absorbs \large \boxed{\textbf{105 kcal}} }$