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Question

Calculate the volume of Cl2(g) (in mL) liberated at 1 atm, 273 K when 1.74 gm MnO2 reacts with 2.19 gm HCl according to the following reaction with 40% yield.
MnO2+HClMnCl2+Cl2+H2O
(Atomic mass of Mn=55,Cl=35.5)

A
336 mL
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B
112 mL
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C
134.4 mL
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D
44.8 mL
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Solution

The correct option is C 134.4 mL
MnO2+4HClMnCl2+Cl2+2H2O
1.7487 2.1936.5
=2 =.06
0.06 mol HCl produces 14×0.06 mol of Cl2
=14×.06×.4=.006 mol (with 40% yield).
Volume of Cl2=nRTP=0.006×22400 =134.4 mL.

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