# Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO 3 ( s ) . The equation for the reaction is 2 KClO 3 ⟶ 2 KCl + 3 O 2 Calculate how many grams of O 2 ( g ) can be produced from heating 82.4 g KClO 3 ( s ) .

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## Question:

Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO 3 ( s ) . The equation for the reaction is 2 KClO 3 ⟶ 2 KCl + 3 O 2 Calculate how many grams of O 2 ( g ) can be produced from heating 82.4 g KClO 3 ( s ) .

Mass of produced = 32 g

Explanation:

Calculation of the moles of as:-

Mass = 82.4 g

Molar mass of = 122.55 g/mol

The formula for the calculation of moles is shown below: Thus,  From the reaction shown below:- 2 moles of potassium chlorate on reaction forms 3 moles of oxygen gas

So,

0.67237 moles of potassium chlorate on reaction forms moles of oxygen gas

Moles of oxygen gas = 1 mole

Molar mass of oxygen gas  = 32 g/mol

Mass of produced = 32 g

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