An excess of sodium hydroxide was added to 250 ml of 32% nitric acid with a density of 1.2 g / ml.

An excess of sodium hydroxide was added to 250 ml of 32% nitric acid with a density of 1.2 g / ml. Determine the mass of the salt formed.

Given:

V (HNO3) = 250 ml

p (HNO3) = 1.2 g / ml

w% (HNO3) = 32%

To find:

m (salt) -?

Solution:

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

1) Find the mass of the nitric acid solution:

m (HNO3) = p * V = 250 ml * 1.2 g / ml = 300 g

2) Find the mass of nitric acid in solution:

m (HNO3) = 300 g * 0.32 = 96 g

3) Find the amount of acid:

n (HNO3) = m: M = 96 g: 63 g / mol = 1.52 mol

4) We compose a logical expression:

if 1 mol of HNO3 gives 1 mol of NaNO3,

then 1.52 mol of HNO3 will give x mol of NaNO3,

then x = 1.52 mol.

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

m (NaNO3) = n * M = 1.52 mol * 85 g / mol = 129.2 g.

Answer: m (NaNO3) = 129.2 g.