What volume of oxygen will be released during the decomposition of AgNO3 weighing 400 g

What volume of oxygen will be released during the decomposition of AgNO3 weighing 400 g, containing a mass fraction of silver nitrate 98%?

To solve the problem, we compose the reaction equation, arrange the coefficients:
2AgNO3 = 2NO2 + O2 + 2Ag – redox reaction, oxygen is released;
M (AgNO3) = 169.8 g / mol; M (O2) = 32 g / mol;
Let us calculate the mass of silver nitrate if the mass fraction of the yield is known:
W = m (practical) / m (theoretical) * 100;
m (theoretical) = 400 / 0.98 = 408.16 g;
Let us determine the number of moles of AgNO3 (theoretical):
Y (AgNO3) = m / M = 408.16 / 169.8 = 2.4 mol;
Let’s make the proportion:
2.4 mol (AgNO3) – X mol (O2);
-2 mol -1 mol hence, X mol (O2) = 1 * 2.4 / 2 = 1.2 mol;
We find the volume of oxygen:
V (O2) = 1.2 * 22.4 = 26.98 liters.
Answer: in the course of the reaction, oxygen with a volume of 26.98 liters was released.