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Equal weights of Zn metal and iodine are...

Equal weights of Zn metal and iodine are mixed together and `I_(1)` is completley converted to `ZnI_(2)`. What fractionn by weight of original Zn remains unreacted? (Zn=65,I=127)

A

0.34

B

0.74

C

0.84

D

Unable to predict

Text Solution

Verified by Experts

The correct Answer is:
B
`Zn+I_(1) rarr ZnI_(2)`
Initially `(x)/(65)` moles `(x)/(254)` moles
Finally `((x)/(65) - (x)/(254))` moles - `(x)/(254)` moles
Fraction by weight of original Zn that remains unreacted `= (((x)/(65)- (x)/(254))xx 65)/(((x)/(65)) xx 65)= 0.74`
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