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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 :

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Answer:

[tex]\large \boxed{\text{105 kcal}}[/tex]

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

[tex]\text{Moles of NO} = \text{147 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{4.898 mol NO}[/tex]

2. Amount of heat

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

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