159.5 g of KClO3 is reacted in a way that has a percent yield of 83 %. What mass of oxygen gas is actually produced?
In the reaction: KClO3 ==> KCl + O2
Find MW of KClO3 = 106.54 g/mol
Find MW of O2 = 32 g/mol
159.5 g KCIO3 X 1 mol KClO3/ 106.54 g KClO3 X 1 mol O2/ 1 mol KClO3 = 1.50 mol O2
1.50 mol O2 X 32 g O2/ 1 mol O2 = 47.91 g O2
Since the actual percent yield is 83%.
Multiply O2 yield by 83%.
47.91 g O2 X 0.83 = 39.76 g O2.
Actual yield is 39.76 g O2.
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