Ammonia, with a volume of 11.2 liters, was passed into a solution containing 35 g of nitric acid

Ammonia, with a volume of 11.2 liters, was passed into a solution containing 35 g of nitric acid. Determine the masses of substances in solution after the end of the reaction.

Let’s implement the solution according to the plan:
1. Let’s write down the reaction equation, arrange the coefficients:
5NH3 + 3HNO3 = 4N2 + 9H2O – redox reaction, nitrogen and water are released;
2. Let’s make calculations using the formulas:
M (HNO3) = 1 + 14 + 16 * 3 = 63 g / mol;
M (N2) = 28 g / mol;
M (H2O) = 18 g / mol;
3. Determine the amount of mol of ammonia:
1 mol of gas at n. y – 22.4 l;
X mol (NH3) -11.2 L from here, X mol (NH3) = 1 * 11.2 / 22.4 = 0.5 mol;
4. Let’s calculate the number of mol of nitric acid by the formula:
Y (HNO3) = m / M = 35/63 = 0.5 mol;
5. Determine the amount of mol of nitrogen:
0.5 mol (HNO3) – X mol (N2);
-3 mol                   -4 mol from here, X mol (N2) = 0.5 * 4/3 = 0.66 mol;
6. Determine the amount of moles of water:
0.5 mol (HNO3) – X mol (H2O);
-3 mol                  – 9 mol from here, X mol (H2O) = 0.5 * 9/3 = 1.5 mol;
7. Let’s calculate the masses of nitrogen and water:
m (N2) = Y * M = 28 * 0.66 = 18.48 g;
m (H2O) = 1.5 * 18 = 27 g.
Answer: the mass of nitrogen is 18.48 g, the mass of water is 27 g.