In which case more oxygen will be obtained: when decomposing 5 g of potassium chlorate

In which case more oxygen will be obtained: when decomposing 5 g of potassium chlorate or when decomposing 5 g of potassium permanganate.

1. Calculate the amount of potassium chlorate and potassium permanganate substance according to the formula:

n = m: M, where M is molar mass.

M (KClO3) = 39 + 35 + 16 × 3 = 122 g / mol.

М (КМnO4) = 39 + 55 + 16 × 4 = 158 g / mol.

n (KClO3) = 5 g: 122 g / mol = 0.04 mol.

n (KMnO4) = 5 g: 158 g / mol = 0.03 mol.

2. Let us compose the equation for the decomposition of potassium chlorate, find the quantitative ratios of potassium chlorate and oxygen.

2KClO3 = 2KCl + 3O2.

Upon decomposition of 2 mol of KClO3, 3 mol of oxygen are formed; they are in quantitative ratios of 2: 3.

2 mol KClO3 – 3 mol (О2),

0.04 mol KClO3 – n mol (О2),

n mol (О2) = (0.04 × 3 mol): 2 mol,

n mol (O2) = 0.06 mol.

3.Let’s find the mass O2 by the formula:

m = n M, M (O2) = 32 g / mol.

m (O2) = 0.06 × 32 = 1.92 g.

4. Let’s compose the reaction equation for the decomposition of potassium permanganate.

2KMnO4 = K2MnO4 + MnO2 + O2.

Let’s find the quantitative ratios of potassium permanganate and oxygen.

When 2 mol of permanganate decomposes, 1 mol of oxygen is formed, that is, the amount of oxygen substance will be 2 times less than the amount of permanganate substance.

n (O2) = 0.03: 2 = 0.015 mol.

2 mol КМnO4 – 1 mol О2,

0.03 mol КМnO4 – n mol О2,

n mol О2 = (0.03 × 1): 2,

n mol O2 = 0.015 mol.

Let’s find the mass of O2.

m (О2) = 0.015 mol × 32 = 0.48 g.

Answer: more oxygen will be obtained by decomposition of potassium chlorate 1.92 g, and by decomposition of potassium permanganate 0.78 g.