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Question

Calculate the weight of MnO2.
HCl of specific gravity 1.2 g/mL and 5% by weight needed to produce 1.12 L of Cl2 at STP by the reaction (as nearest integer).
MnO2+4HClMnCl2+3H2O+Cl2

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Solution

(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=60.97 mL
Because HCl is also used to give MnCl2 thus volume used is double than required for the reduction of MnO2.
VHCl=2×60.97=121.94 mL
(c) Also, mEq of MnO2=mEq of HCl=100
W87/2×103=100(Ew of MnO2=55+322)
WMnO2=4.35 g
nearest integer value is 4 gm.

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