How many moles of oxygen are formed during the decomposition of sodium nitrate with a mass of 240.6 g.

To solve, we write down the process equation and data by the condition:

m = 240.6 g Y -?

2NaNO3 = O2 + 2NaNO2 – decomposition, oxygen is released, sodium nitrite.
We make calculations using the formulas:
M (NaNO3) = 84.9 g / mol;

M (O2) = 32 g / mol;

Y (NaNO3) = m / M = 240.6 / 84.9 = 2.8 mol.

Proportion:
2.8 mol (NaNO3) – X mol (O2);

-2 mol -1 mol hence, X mol (O2) = 2.8 * 1/2 = 1.4 mol.

Answer: oxygen was obtained in the amount of 1.4 mol.