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Question

If x g of MnO2 and the volume of HCl of specific gravity 1.2g mL3 and 5% by weight needed to produce 1.12L of Cl2 at STP by the reaction,
MnO2+4HClMnCl2+3H2O+Cl2
then value of 100X is :

A
435
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B
450
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C
400
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D
none of these
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Solution

The correct option is A 435
(a) NHCl=% by weight×10×dEwofHCl
=5×10×1.236.5=1.64
Now, mEq of MnO2= mEq of HCl
=mEq of Cl2 formed.
=1.1211.2×103
=100 EwofCl2=M21eqofCl2=11.2L
(b) Volume of HCl used:
N×V=100
1.64×V=100
VHCl=60.97mL
Because HCl is also used to give MnCl2 thus volume used is double than required for the reduction of MnO2.
VHCl=2×60.97=121.94mL
(c) Also, mEq of MnO2=mEq of HCl=100
W87/2×103=100(Ew of MnO2=55+322)
WMnO2=4.35g

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