# What volume of a 60% nitric acid solution with a density of 1.305 g / ml can be obtained using the nitrogen-containing

What volume of a 60% nitric acid solution with a density of 1.305 g / ml can be obtained using the nitrogen-containing product of the catalytic oxidation of 896 liters of ammonia?

1) Let’s compose the reaction equations:
4NH3 + 5O2 = 4NO ↑ + 6H2O
2NO + O2 = 2NO2 ↑
4NO2 + O2 + H2O = 4HNO3
The process does not generate nitrogen-containing by-products, i.e. all nitrogen atoms included in NH3 go into HNO3. Therefore, you can draw up a scheme for calculating:
NH3… → HNO3
2) Calculate the amount of ammonia substance:
n (NH3) = V (NH3) / Vm = 896 / 22.4 = 40 mol
3) Calculate the mass of the HNO3 solution:
According to the calculation scheme:
n (HNO3) = n (NH3) = 40 mol
m (HNO3) = n (HNO3) ∙ M (HNO3) = 40 ∙ 63 = 2520 g.
m (solution) = m (HNO3) ∙ 100% / (ω (HNO3) = 2520 ∙ 60/100 = 4200 g.
4) Calculate the volume of the HNO3 solution:
V = m / ρ
V (HNO3) = m (solution) / ρ (HNO3) ∙ = 4200 / 1.305 = 3218.4 ml = 3.22 liters.