Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO3(s)KClO3(s) . The equation for the reaction is   2KClO3⟶2KCl+3O22KClO3⟶2KCl+3O2   Calculate how many grams of O2(g)O2(g) can be produced from heating 68.0 g KClO3(s)68.0 g KClO3(s) . STRATEGY: Convert the mass of KClO3KClO3 to moles. Convert the number of moles of KClO3KClO3 to the number of moles of O2O2 . Convert the number of moles of O2O2 to grams. Step 1: 68.0 g KClO368.0 g KClO3 is equal to 0.555 mol KClO30.555 mol KClO3 . Step 2: If 0.555 mol KClO30.555 mol KClO3 reacted, how much O2O2 was produced?

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Q: to

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Stoichiometry

Stoichiometry is a chemical division that evaluates the quantitative content of the chemical process with the help of the reacting molecules or products taking part in the reaction. The term “stoicheion” in “Stoichiometry”, is an old Greek term for "element" whereas “metron” means "measure".