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 ) .

Answer:

Answer:

Mass of ?f=O 2 produced = 32 g

Explanation:

Calculation of the moles of ?f=KClO 3 as:-

Mass = 82.4 g

Molar mass of ?f=KClO 3 = 122.55 g/mol

The formula for the calculation of moles is shown below:

?f=moles%20%3D%20%5Cfrac%7BMass%5C%20taken%7D%7BMolar%5C%20mass%7D

Thus,

?f=Moles%3D%20%5Cfrac%7B82.4%5C%20g%7D%7B122

?f=Moles%3D%200

From the reaction shown below:-

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?f=2KClO 3%5Crightarrow%202KCl%2B3O 2

2 moles of potassium chlorate on reaction forms 3 moles of oxygen gas

So,

0.67237 moles of potassium chlorate on reaction forms ?f=%5Cfrac%7B3%7D%7B2%7D%5Ctimes%200 moles of oxygen gas

Moles of oxygen gas = 1 mole

Molar mass of oxygen gas  = 32 g/mol

Mass of ?f=O 2 produced = 32 g

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