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An excess of hydrogen gas reacts with 42.5 g of nitrogen gas. What volume (in liters) of ammonia is produced at STP?H2   + N2   →   NH3

Question

An excess of hydrogen gas reacts with 42.5 g of nitrogen gas. What volume (in liters) of ammonia is produced at STP?

H2+N2NH3 \text{H}_2 + \text{N}_2 \rightarrow \text{NH}_3

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Solution

Sure, here are the steps to solve this problem:

Step 1: Balance the chemical equation. The balanced chemical equation is: N2 + 3H2 → 2NH3

Step 2: Convert the mass of nitrogen gas to moles. The molar mass of nitrogen (N2) is approximately 28 g/mol. So, 42.5 g of nitrogen gas is about 1.52 moles (42.5 g ÷ 28 g/mol = 1.52 mol).

Step 3: Use the stoichiometry of the reaction to find the moles of ammonia. From the balanced chemical equation, we can see that 1 mole of nitrogen gas (N2) produces 2 moles of ammonia (NH3). So, 1.52 moles of nitrogen gas will produce about 3.04 moles of ammonia (1.52 mol × 2 = 3.04 mol).

Step 4: Convert the moles of ammonia to volume. At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. So, 3.04 moles of ammonia will occupy about 68.1 liters (3.04 mol × 22.4 L/mol = 68.1 L).

So, the volume of ammonia produced at STP is 68.1 liters.

This problem has been solved

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