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Hyrdrogen gas and nitrogen gas react in the presence of a catylist to produce ammonia gas. at SATP, 71.8 l of nitrogen gas and 169.4 l of hydrogen gas react, calculate the mass (in grams) of ammonia gas produced.
Accepted Answer
At SATP, 1 mole of any gas occupies 22.4 L. Therefore, 71.8 L of nitrogen gas is equal to 71.8 L / 22.4 L/mol = 3.2 mol of nitrogen gas. Similarly, 169.4 L of hydrogen gas is equal to 169.4 L / 22.4 L/mol = 7.56 mol of hydrogen gas. The balanced chemical equation for the reaction is:
N2(g) + 3H2(g) → 2NH3(g)
From the balanced chemical equation, we can see that 1 mole of nitrogen gas reacts with 3 moles of hydrogen gas to produce 2 moles of ammonia gas. Therefore, 3.2 mol of nitrogen gas will react with 3.2 mol * 3 = 9.6 mol of hydrogen gas to produce 3.2 mol * 2 = 6.4 mol of ammonia gas. The molar mass of ammonia gas is 17.03 g/mol. Therefore, 6.4 mol of ammonia gas has a mass of 6.4 mol * 17.03 g/mol = 109.1 g.
Therefore, the mass of ammonia gas produced is 109.1 g.
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