The following equation is one way to prepare oxygen in a lab. 2KClO3 → 2KCl + 3O2 Molar mass Info: MM O2 = 32 g/mol MM KCl = 74.55 g/mol MM KClO3 = 122.55 g/mol If 4.00 moles of KClO3 are totally consumed, how many grams of oxygen gas would be produced? 192 g 6.00 g 85.3 g 735 g

Question

Janiya Nash

Chemistry

3 November, 14:23

The following equation is one way to prepare oxygen in a lab. 2KClO3 → 2KCl + 3O2 Molar mass Info: MM O2 = 32 g/mol MM KCl = 74.55 g/mol MM KClO3 = 122.55 g/mol If 4.00 moles of KClO3 are totally consumed, how many grams of oxygen gas would be produced? 192 g 6.00 g 85.3 g 735 g

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Asa Grimes · Answer 1

From the equation above the reacting ratio of KClO3 to O2 is 2:3 therefore the number of moles of oxygen produced is (4 x3) / 2 = 6 moles since four moles of KClO3 was consumed

mass=relative formula mass x number of moles

That is 32g/mol x 6 moles = 192grams