# The gas that is released during the decomposition of 1032 g of potassium chlorate containing 5%

The gas that is released during the decomposition of 1032 g of potassium chlorate containing 5% oxygen-free impurities was used to oxidize sulfur. Determine the mass of sulfur that can be reacted and the volume of the resulting gaseous oxidation product.

Let us find the mass of potassium chlorate without impurities.

100% – 5% = 95%.

1032 g – 100%,

X – 95%.

X = (1032 g × 95%): 100% = 980.4 g.

Find the amount of potassium chlorate substance by the formula:

n = m: M.

M (KClO3) = 122 g / mol.

n = 980.4 g: 122 g / mol = 8.036 mol.

Let’s compose the reaction equation, find the quantitative ratios of substances.

2 KClO3 → 2KCl + 3O2 ↑.

According to the reaction equation, there is 3 mol of oxygen for 2 mol of potassium chlorate. Substances are in quantitative ratios of 2: 3.

The amount of oxygen will be 1.5 times greater than the amount of potassium chlorate.

n (О2) = 1.5n (КClO3) = 8.036 × 1.5 = 12.054 mol.

2 mol КClO3 – 3 mol O2,

8.036 mol KClO3 – n mol O2.

n mol O2 = (8.036 × 3): 2 = 12.054 mol.

Let’s find the mass of oxygen by the formula:

m = n × M,

M (O2) = 32 g / mol.

m = 12.054 mol × 32 g / mol = 385.73 g.

Let’s find the quantitative ratios of substances.

S + O2 = SO2.

For 1 mole of sulfur there is 1 mole of oxygen, 1 mole of sulfur oxide.

Substances are in quantitative ratios 1: 1: 1.

The amount of substance will be the same.

n (S) = n (O2) = n (SO2) = 12.054 mol.

Let’s find the mass of sulfur.

m = n × M.

M (S) = 32 g / mol.

m = 32 g / mol × 12.054 mol = 385.73 g.

Find the volume of sulfur oxide.

V = Vn n, where Vn is the molar volume of gas, equal to 22.4 l / mol, and n is the amount of substance.

V = 22.4 L / mol × 12.054 mol = 270 L.