#### Question

# Moles and Mass Question!?

Ammonia is manufactured from the reaction between nitrogen and hydrogen:

N2 (g) + 3 H2 (g) ? 2 NH3 (g)

What is the maximum mass of ammonia that can be obtained from a mixture of 56 g of nitrogen with 9 g of hydrogen? (Relative atomic masses: H = 1, N = 14)

#### Answers

68.12 g

Work:

1. Convert grams into moles. 56/14=4 mol N2 9/1= 9 mol of H2

2. Find limiting reagent based on formula ratios. 4mol N2* (2mol NH3/1 mol of N2)= 8 mol

9 mol H2* (2 mol NH3/3 mol H2)= 6 mol. Therefore H2 is your limiting reagent.

3. Calculate Theoretical Yield. 6 mol H2 * (2 mol NH3/3 mol H2) * (17.03 g/1 mol NH3)=68.12

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