What mass of potassium permanganate is needed to obtain 96 g of oxygen.

Let’s implement the solution:

According to the condition of the problem, we compose the equation of the process:
Xr -? m = 96 g

2KMnO4 = K2MnO4 + MnO2 + O2 – decomposition, oxygen is released;

Let’s make calculations using the formulas of substances:
M (KMnO4) = 158 g / mol;

M (O2) = 32 g / mol;

Y (O2) = m / M = 96/32 = 3 mol.

Proportion:
X mol (KMnO4) – 3 mol (О2);

-2 mol 1 mol from here, X mol (KMnO4) = 2 * 3/1 = 6 mol.

We find the mass of the original substance:
m (KMnO4) = Y * M = 6 * 158 = 948 g

Answer: the mass of potassium permanganate is 948 g