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Question

Cl2+2KOHKCl+KClO+H2O
3KClO2KCl+KClO3
4KClO3KCl+3KClO4

2 moles of chlorine gas reacted with an excess of potassium hydroxide and 50 g of potassium perchlorate was obtained. Calculate the percentage yield of perchlorate in the reaction.
Molar mass of potassium perchlorate = 138.5 g/mol

A
72 %
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B
60 %
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C
90 %
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D
49 %
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Solution

The correct option is A 72 %
The complete balanced equation is:
12Cl2+24KOH21KCl+3KClO4+12H2O

12 moles of chlorine gives 3 moles of perchlorate.
2 moles of chlorine gives 0.5 moles of perchlorate.
Theoretical yield of perchlorate = 0.5 mole = 69.25 g
Experimental yield of perchlorate = 50 g
Percentage yield of perchlorate = experimental yieldtheoretical yield×100 = 50×10069.25=72%

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