Show that the following equation obeys the Law of Conservation of Mass. The amount of the products equals the amount of the reactant. You begin with 247 grams of KClO3 2KClO3  2KCl + 3O2

Answer 1

2KClO₃ ==> 2KCl + 3O₂ ... balanced equation for decomposition of KClO₃

moles of KClO₃ started with = 247 g KClO3 x 1 mol KClO3/123 g = 2.01 moles KClO₃

moles KCl produced = 2.01 mol KClO₃ x 2 mol KCl/2 mol KClO₃ = 2.01 moles KCl produced

moles O₂ produced = 2.01 mol KClO₃ x 3 mol O₂/2 mol KClO₃ = 3.02 moles O₂ produced

Now we convert moles of products to grams of products and compare that to grams of starting material.

grams KCl = 2.01 mol KCl x 74.6 g/mole = 150 g

grams O₂ = 3.02 moles O₂ x 32 g/mol = 96.6 g

Total mass of products = 150 g + 96.6 g = 246.6 g = 247 g of products = original mass of starting material