Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO3 (s). The equation for the reaction is

2KClO3 - - - > 2KCl+3O2

Calculate how many grams of O2 (g) can be produced from heating 72.0 grams of KClO3 (s).

Question

Alisha

Chemistry

3 March, 14:12

Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO3 (s). The equation for the reaction is

2KClO3 - - - > 2KCl+3O2

Calculate how many grams of O2 (g) can be produced from heating 72.0 grams of KClO3 (s).

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Answers (1)

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Miriam David · Answer 1

2 KClO3 → 2 KCl + 3 O2

(19.7 g KClO3 / (122.5495 g KClO3/mol) x (3 mol O2 / 2 mol KClO3) x (31.99886 g O2/mol) = 7.72 g O2

I hope this is correct. Boi, this is so confusing

Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO3 (s). The equation for the reaction is2KClO3 - - - > 2KCl+3O2Calculate how many grams of O2 (g) can be produced from heating 72.0 grams of KClO3 (s).

Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO3 (s). The equation for the reaction is

2KClO3 - - - > 2KCl+3O2

Calculate how many grams of O2 (g) can be produced from heating 72.0 grams of KClO3 (s).