How much nitrogen and hydrogen must react to create 20 g of ammonium?

Let’s write down the reaction equation, arrange the coefficients:
N2 + 3H2 = 2NH3 – reaction of the compound, ammonia was obtained;
Let’s define the molar masses:
M (N2) = 14 * 2 = 28 g / mol;
M (H2) = 2 g / mol;
M (NH3) = 14 + 3 = 17 g / mol;
Let’s calculate the amount of moles of ammonia, nitrogen, hydrogen:
Y (NH3) = m / M = 20/17 = 1.176 mol;
X mol (N2) – 1.176 mol (NH3);
-1 mol -2 mol hence, X mol (N2) = 1 * 1.176 / 2 = 0.588 mol;
X mol (H2) – 1.176 mol (NH3);
-3 mol -2 mol hence, X mol (H2) = 3 * 1.176 / 2 = 1.764 mol;
We find the volume of nitrogen, hydrogen:
Y (N2) = 0.558 * 22.4 = 12.49 L;
Y (H2) = 1.764 * 22.4 = 39.5 liters.
Answer: the volume of nitrogen is 12.49 liters, the volume of hydrogen is 39.5 liters.