# 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.

Answer: 3.22 L