What volume of hydrogen, reacting with nitrogen, will produce 500 dm^3 of NH3 at STP?

Answer 1

To solve this problem, we will use the stoichiometry of the reaction and the ideal gas law.

The balanced chemical equation for the reaction between hydrogen gas (H2) and nitrogen gas (N2) to produce ammonia gas (NH3) is:

N2 (g) + 3H2 (g) → 2NH3 (g)

From the balanced equation, we can see that for every molecule of N2, 3 molecules of H2 are required, and 2 molecules of NH3 are produced. Since all the gases are at STP, we can use the molar volume of a gas at STP, which is 22.4 L/mol or 22.4 dm³/mol.

First, calculate the number of moles of NH3 produced:

volume of NH3 = 500 dm³;
molar volume of a gas at STP = 22.4 dm³/mol;

n(NH3) = volume of NH3 / molar volume = 500 dm³ / 22.4 dm³/mol = 22.32 mol of NH3

Now, use the stoichiometry to calculate the number of moles of H2:

From the balanced equation: 2 moles of NH3 are produced from 3 moles of H2

n(H2) = n(NH3) * (3 mol H2 / 2 mol NH3) = 22.32 mol * (3/2) = 33.48 mol of H2

Finally, find the volume of H2 at STP:

volume of H2 = n(H2) * molar volume = 33.48 mol * 22.4 dm³/mol = 749.95 dm³

So, approximately 750 dm³ of hydrogen gas is required to produce 500 dm³ of ammonia gas at STP.