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?

Question
Asked Nov 6, 2019
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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:

  1. Convert the mass of KClO3KClO3 to moles.

  2. Convert the number of moles of KClO3KClO3 to the number of moles of O2O2 .

  3. 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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Expert Answer

Step 1

The moles of KClO3 can be calculated as,

The...

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2KCIO 2KC302 1 mol KC1O3 39.1 +35.5 3(16.0) 122.6 g 1 mol O2 2(16.0) =32.0 g

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