Answer to Question #340355 in Physical Chemistry for Tea

Question

The reaction of ammonia gas to nitrogen monoxide gas yields nitrogen gas and water vapor. How many moles of each reactant will be present if 13.8 moles of nitrogen gas is produced?

Answer 1

Answer:

n NH3 = 10.96 mol

n NO = 16.44 mol

Explanation:

Step 1: Write the balanced chemical equation.

4NH3 + 6NO → 5N2 + 6H2O

Step 2: Compute for the number of moles of each reactant present.

For mol of NH3

n NH3 = 13.7 mol N2 × (4 mol NH3 / 5 mol N2)

n NH3 = 10.96 mol

For mol of NO

n NO = 13.7 mol N2 × (6 mol NO / 5 mol N2)

n NO = 16.44 mol

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