What volume of oxygen will be released during the decomposition of 510 grams of technical

What volume of oxygen will be released during the decomposition of 510 grams of technical sodium nitrate with 20 percent of impurities?

1. Let’s write down the reaction of sodium nitrate decomposition:

2NaNO3 = 2NaNO2 + O2;

2.Calculate the mass of pure sodium nitrate:

m (NaNO3) = mtechn (NaNO3) – m (impurities);

m (impurities) = w (impurities) * mtechn (NaNO3) = 0.2 * 510 = 102 g;

m (NaNO3) = 510-102 = 408 g;

3. find the chemical amounts of nitrate and oxygen:

n (NaNO3) = m (NaNO3): M (NaNO3);

M (NaNO3) = 23 + 14 + 3 * 16 = 85 g / mol;

n (NaNO3) = 408: 85 = 4.8 mol;

n (O2) = n (NaNO3) = 4.8 mol;

4.Calculate the volume of released oxygen:

V (O2) = n (O2) * Vm = 4.8 * 22.4 = 107.52 dm3.

Answer: 107.52 dm3.