Question

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

N2(g)+O2(g)→2NO(g), ΔH = + 43 kcal/molHow much heat is absorbed during production of 147 g of NO by the combination of nitrogen and oxygen?
N2(g)+O2(g)→2NO(g), ΔH = + 43 kcal/mol

Answer 1

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

Step-by-step 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} * \frac{\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 } * \frac{\text{43 kcal}}{\text{2 mol NO}} = \text{105 kcal}\\\\\text{The reaction absorbs $\large \boxed{\textbf{105 kcal}}$}`