When 17 g of sodium nitrate were reacted with concentrated sulfuric acid, 11.5 g of nitric acid were obtained.

When 17 g of sodium nitrate were reacted with concentrated sulfuric acid, 11.5 g of nitric acid were obtained. Determine the mass fraction of the yield of nitric acid from the theoretical yield?

Let’s implement the solution:
1. We write down the process according to the condition:
m = 17 g. m = 11.5 g; W -?
2NaNO3 + H2SO4 = Na2SO4 + 2HNO3 – ion exchange, nitric acid obtained;
2. Let’s make calculations using the formulas:
M (NaNO3) = 84.9 g / mol.
M (HNO3) = 63 g / mol.
Y (NaNO3) = m / M = 17 / 84.9 = 0.2 mol.
Y (HNO3) = 0.2 mol since the amount of substances according to the equation is 2 mol.
3. Find the mass of the product, as well as the yield in percent:
m (HNO3) = Y * M = 0.2 * 63 = 12.6 g (theoretical weight).
W (HNO3) = 11.5 / 12.6 * 100 = 91.27%
Answer: obtained nitric acid weighing 12.6 g, the yield was 91.27%.