# 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. Remember: The process of learning a person lasts a lifetime. The value of the same knowledge for different people may be different, it is determined by their individual characteristics and needs. Therefore, knowledge is always needed at any age and position.