# 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.

Answer: 270 l; 385.73 g.