Calculate m NaNO3 formed by the interaction of 630 kg of a solution in which the mass fraction of w HNO3

Calculate m NaNO3 formed by the interaction of 630 kg of a solution in which the mass fraction of w HNO3 is 50% from 170 kg of a solution containing NaOH with a mass fraction of w = 40%

Given:

m (HNO3) = 630 kg

m (NaOH) = 170 kg

w% (HNO3) = 50%

w% (NaOH) = 40%

To find:

m (NaNO3) -?

Solution:

HNO3 + NaOH = NaNO3 + H2O, – we solve the problem, relying on the composed reaction equation:

1) Find the masses of sodium hydroxide and nitric acid in solutions:

m (HNO3) = 630 kg * 0.5 = 315 kg

m (NaOH) = 170 kg * 0.4 = 68 kg

2) Find the amount of alkali and acid:

n (HNO3) = m: M = 315 kg: 63 kg / kmol = 5 kmol

n (NaOH) = m: M = 68 kg: 40 kg / kmol = 1.7 kmol

We start from a lower value to make the calculations more accurate. We work with NaOH:

3) We compose a logical expression:

if 1 kmol of NaOH gives 1 kmol of NaNO3,

then 1.7 kmol NaOH will give x kmol NaNO3,

then x = 1.7 kmol.

4) Find the mass of sodium nitrate formed during the reaction:

m (NaNO3) = n * M = 1.7 kmol * 85 kg / kmol = 144.5 kg.

Answer: m (NaNO3) = 144.5 kg.