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+HCl→MnCl2+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 C134.4 mL MnO2+4HCl→MnCl2+Cl2+2H2O 1.74872.1936.5 =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.