If x g of MnO2 and the volume of HCl of specific gravity 1.2gmL−3 and 5% by weight needed to produce 1.12L of Cl2 at STP by the reaction, MnO2+4HCl→MnCl2+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=%byweight×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.