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Question

# The minimum weight of MnO2 and minimum volume of HCl of specific gravity 1.2 and 3.65% by weight, needed to produce 1.12 litres of Cl2 at STP by the reaction given as follows:MnO2+HCl→MnCl2+H2O+Cl2

A
4.35g,200ml
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B
48.7g,166.67ml
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C
4.35g,166.7ml
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D
4.35g,333.3ml
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Solution

## The correct option is B 4.35g,166.7mlThe number of moles of chlorine obtained =1.1222.4=0.05 moles.They are also equal to the number of moles of manganese dioxide.Mass of manganese dioxide =0.05×86.9=4.35 g.The number of moles of HCl =4×0.05=0.2 moles.Mass of HCl =0.2×36.5×1003.65=200 g.Volume of HCl =2001.2=166.7 ml.

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