From 34 g of sodium nitrate by the action of concentrated sulfuric acid, 22.7 g

From 34 g of sodium nitrate by the action of concentrated sulfuric acid, 22.7 g of nitric acid were obtained. What is the nitrogen yield (in%) of the theoretically possible.

Given:

m (NaNO3) = 34 g

m (HNO3) = 22.7 g

To find:

w% (HNO3 practical) -?

Decision:

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

1) Find the amount of sodium nitrate that has reacted:

n (NaNO3) = m: M = 34 g: 85 g / mol = 0.4 mol

2) We compose a logical expression:

if 2 mol NaNO3 gives 2 mol HNO3,

then 0.4 mol NaNO3 will give x mol HNO3,

then x = 0.4 mol.

3) Find the mass of nitric acid:

m (HNO3) = n * M = 0.4 mol * 63 g / mol = 25.2 g.

4) Find the practical yield of nitric acid:

w% (HNO3 practical) = 22.7 g: 25.2 g * 100% = 90%.

Answer: w% (HNO3 practical) = 90%.