Calculate the mass of a nitric acid solution with a mass fraction of a substance of 0.2, which is necessary

Calculate the mass of a nitric acid solution with a mass fraction of a substance of 0.2, which is necessary for the complete nitrolization of a sodium hydroxide solution in a mass of 120 grams, with a mass fraction of alkali of 0.16.

Given:

m (NaOH) = 120 g

w% (NaOH) = 16%

w% (HNO3) = 20%

To find:

m (solution HNO3) -?

Decision:

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

1) Find the mass of sodium hydroxide in the solution:

m (NaOH) = 120 g * 0.16 = 19.2 g

2) Find the amount of sodium hydroxide that has reacted:

n (NaOH) = m: M = 19.2 g: 40 g / mol = 0.48 mol

3) We compose a logical expression:

If 1 mole of NaOH requires 1 mole of HNO3,

then 0.48 mol NaOH will require x mol HNO3,

then x = 0.48 mol.

4) Find the mass of nitric acid:

m (HNO3) = n * M = 0.48 mol * 63 g / mol = 30.24 g

5) Find the mass of the nitric acid solution:

m (solution HNO3) = 30.24 g: 0.2 = 151.2 g

Answer: m (solution HNO3) = 151.2 g.