How many grams of KCl and oxygen are formed during the decomposition of 1.22 g of KClO3-mole berthollet salt.

Given:
m (KClO3) = 1.22 g

To find:
m (KCl) -?
m (O2) -?

Decision:
1) Write the reaction equation:
2KClO3 => 2KCl + 3O2;
2) Calculate the amount of substance KClO3:
n (KClO3) = m (KClO3) / M (KClO3) = 1.22 / 122.5 = 0.01 mol;
3) Determine the amount of KCl substance:
n (KCl) = n (KClO3) = 0.01 mol;
4) Calculate the mass of KCl:
m (KCl) = n (KCl) * M (KCl) = 0.01 * 74.5 = 0.745 g;
5) Determine the amount of substance O2:
n (O2) = n (KClO3) * 3/2 = 0.01 * 3/2 = 0.015 mol;
6) Calculate the mass of O2:
m (O2) = n (O2) * M (O2) = 0.015 * 32 = 0.48 g.

Answer: The mass of KCl is 0.75 g; O2 – 0.48 g.