When 68 g of sodium nitrate was treated with an excess of concentrated sulfuric acid, nitric acid weighing 45.36 g

When 68 g of sodium nitrate was treated with an excess of concentrated sulfuric acid, nitric acid weighing 45.36 g was obtained. Find the fraction of the yield of the reaction product.

Given:
m (NaNO3) = 68 g
m pract. (HNO3) = 45.36 g

To find:
η (HNO3) -?

1) NaNO3 + H2SO4 => HNO3 + NaHSO4;
2) n (NaNO3) = m / M = 68/85 = 0.8 mol;
3) n theory. (HNO3) = n (NaNO3) = 0.8 mol;
4) m theor. (HNO3) = n * M = 0.8 * 63 = 50.4 g;
5) η (HNO3) = m practical. * 100% / m theory. = 45.36 * 100% / 50.4 = 90%.

Answer: The HNO3 yield is 90%.