By decomposition of 5 grams of potassium chlorate, oxygen was obtained. Determine the mass and volume of oxygen.

1. We write down the equation according to the problem statement:
3KClO3 = 3O2 + 2KCl – decomposition, oxygen is formed;
2. Calculations:
M (KClO3) = 122.6 g / mol;
M (O2) = 32 g / mol;
Y (KClO3) = m / M = 5 / 122.6 = 0.04 mol;
Y (O2) = 0.04 mol since the amount of these substances is 3 mol.
3. Find the mass, volume of O2:
m (O2) = Y * M = 0.04 * 32 = 1.28 g;
V (O2) = 0.04 * 22.4 = 0.896 l.
Answer: oxygen was released with a volume of 0.896 liters, weighing 1.28 g.