What mass of salt is formed by the interaction of sodium carbonate 20 g mass containing 20% impurities with nitric acid?

1.Let’s find the mass of pure limestone Na2CO3.
100% – 20% = 80%.
20 g – 100%,
X – 80%,
X = (80% × 20g): 100% = 16g.
2.Let’s find the amount of substance Na2CO3.
n = m: M.
M (Na2CO3) = 106 g / mol.
n = 16 g: 106 g / mol = 0.151 mol.
3. Let’s find the quantitative ratios of substances.
Na2CO3 + 2HNO3 → 2NaNO3 + CO2 + H2O.
For 1 mol of Na2CO3, there are 2 mol of NaNO3.
The substances are in quantitative ratios of 1: 2.
The amount of NaNO3 substance will be 2 times more than the amount of Na2CO3 substance,
n (NaNO3) = 2 n (Na2CO3) = 2 × 0.151 = 0.302 mol.
Let’s find the mass of salt.
m = n × M.
M (NaNO3) = 85 g / mol.
m = 85 g / mol × 0.302 mol = 25.67 g.
Answer: 25.67 g.