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In the following reaction, MnO(2) + 4HC...

In the following reaction, `MnO_(2) + 4HCl rarr MnCl_(2) + 2H_(2)O + Cl_(2)`
2 mole `MnO_(2)` reacts with 4 mol of HCl to form 11.2 L `Cl_(2)` at STP. Thus, percentage yield of `Cl_(2)` is:

A

25%

B

`50%`

C

`100%`

D

75%

Text Solution

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The correct Answer is:
B
`{:(,MnO_(2),+,4HCl,rarr,MnCl_(2),+,2H_(2)O,+,Cl_(2)),("Initially","2 moles",,"4 moles",,-,,-,,-),("Finally","1 mole",,-,,"1 mole",,"2 mole",,"1 mole" = 22.4 L):}`
% yield of `Cl_(2) = (11.2)/(22.4) xx 100% = 50%`
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