What mass of potassium permanganate is required to obtain 11.2 liters of oxygen?

Given:
V (O2) = 11.2 l

To find:
m (KMnO4) -?

Solution:
1) Write the reaction equation:
2 KMnO4 = (tOC) => K2MnO4 + MnO2 + O2;
2) Calculate the molar mass of KMnO4:
M (KMnO4) = Mr (KMnO4) = Ar (K) + Ar (Mn) + Ar (O) * 4 = 39 + 55 + 16 * 4 = 158 g / mol;
3) Calculate the amount of substance O2:
n (O2) = V (O2) / Vm = 11.2 / 22.4 = 0.5 mol;
4) Determine the amount of KMnO4 substance:
n (KMnO4) = n (O2) * 2 = 0.5 * 2 = 1 mol;
5) Calculate the mass of KMnO4:
m (KMnO4) = n (KMnO4) * M (KMnO4) = 1 * 158 = 158 g.

Answer: The mass of KMnO4 is 158 g.