Calculate the volume of oxygen that is formed during the decomposition of 8.8 g of silver nitrate (I).

Let’s write down the data on the condition of the problem:

2AgNO3 = O2 + 2Ag + 2NO2 – decomposition of silver nitrate, oxygen is released;

Calculations:
M (AgNO3) = 169.8 g / mol;

M (O2) = 32 g / mol;

Y (AgNO3) = m / M = 8.8 / 169.0.05 mol.

Proportion:
0.05 mol (Ag (NO3) – X mol (O2);

-2 mol -1 mol from here, X mol (O2) = 0.05 * 1/2 = 0.025 mol.

Find the volume of the product:
V (O2) = 0.025 * 22.4 = 0.58 L

Answer: received oxygen in a volume of 0.58 liters