What mass of nitric acid can be obtained from 170 g of sodium nitrate with concentrated sulfuric acid at a yield of 80% of the acid?

Let’s implement the solution:
1. Let’s compose an equation according to the condition of the problem:
2NaNO3 + H2SO4 = 2HNO3 + Na2SO4 – ion exchange reaction, obtained nitric acid and sodium sulfate;
2. Let’s make the calculations:
M (NaNO3) = 84.9 g / mol;
M (HNO3) = 63 g / mol.
3. Determine the amount of mol of sodium nitrate salt, nitric acid:
Y (NaNO3) = v / m = 170 / 84.9 = 2 mol;
Y (HNO3) = 2 mol, since according to the equation the amount of these substances is 2 mol.
4. Find the theoretical mass of HNO3:
m (HNO3) = Y * M = 2 * 63 = 126 g.
5. The practical weight for nitric acid is calculated by the formula:
W = m (practical) / m (theoretical) * 100;
m (practical) = 0.80 * 126 = 100.8 g.
Answer: the mass of НNO3 was 100.8 g.