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 + N2 → NH3
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.
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