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Question

At STP 89.6 L of chlorine gas reacts with 560 g of KOH according to the following sequential reaction. Calculate the number of grams of KClO4. Molar mass of KClO4=138.5 g mol1
Molar mass of KOH=56 g mol1

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

A
69.25 g
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B
138.5 g
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C
60 g
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D
90 g
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Solution

The correct option is B 138.5 g
22.4 L of chlorine gas = 1 mole of chlorine gas
Given: 89.6 L of chlorine gas at STP
Number of moles of Cl2=89.622.4=4
Number of moles of KOH = 56056= 10 mol
Hence for limiting reagent,
Cl2 = 41
KOH = 102=5
Thus Cl2 is the limiting reagent here.

Final balanced equation
12Cl2+24KOH21KCl+3KClO4+12H2O

3 moles KClO4 is formed from 12 moles of Cl2 and hence number of moles of KClO4 formed from 4 moles of Cl2 =312×4=1
Number of grams of KClO4=1×138.5=138.5 g

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