Calculate the mass of nitric acid formed by the interaction of 86.7 g of technical sodium

Calculate the mass of nitric acid formed by the interaction of 86.7 g of technical sodium nitrate (mass fraction of impurities 2%) with sulfuric acid.

Given:

m (tech. NaNO3) = 86.7 g

w% (impurities) = 2%

Find:

m (HNO3) -?

Solution:

2NaNO3 + H2SO4 = Na2SO4 + 2HNO3, – we solve the problem, relying on the composed reaction equation:

1) Find the mass of pure sodium nitrate:

If the proportion of impurities is 2%, then the proportion of pure substance is 98%

m (NaNO3) = 86.7 g * 0.98 = 85 g

2) Find the amount of sodium nitrate:

n (NaNO3) = m: M = 85 g: 85 g / mol = 1 mol

3) We compose a logical expression:

If 2 mol NaNO3 gives 2 mol HNO3,

then 1 mole of NaNO3 will give x mole of HNO3,

then x = 1 mol.

4) Find the mass of nitric acid formed during the reaction:

m (HNO3) = n * M = 1 mol * 63 g / mol = 63 g.

Answer: m (HNO3) = 63 g.