When sodium nitrate was heated, oxygen with a volume of 280 ml was formed. What mass of salt was decomposed?

Let’s execute the solution:

In accordance with the condition of the problem, we compose the process equation:
2NaNO3 = O2 + 2NaNO2 – OBP, oxygen is evolved;

Calculations:
M (NaNO3) = 84.9 g / mol;

M (O2) = 32 g / mol;

V (O2) = 0.28 l.

Proportions:
1 mol of gas at normal level – 22.4 l

X mol (O2) – 0.28 L from here, X mol (O2) = 1 * 0.28 / 22.4 = 0.01 mol.

X mol (NaNO3) – 0.01 mol (O2);

-2 mol -1 mol hence, X mol (NaNO3) = 2 * 0.01 / 1 = 0.02 mol.

We find the mass of the original substance:
m (NaNO3) = Y * M = 0.02 * 84.9 = 1.7 g

Answer: you need sodium nitrate weighing 1.7 g