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)

Options

A. 0.6
B. 0.74
C. 0.47
D. 0.17

Answer

B

Explanation

Let x g be the initial weight of the Zn metal and iodine each. Since `I_(2)` is completely converted to `ZnI_(2)`, we have, `Zn +I_(2) rarr ZnI_(2)`
Initial no of moles `(x)/(65) (x)/(254) 0`
No of moles at the end of the reaction:
`((x)/(65)-(x)/(254)) 0 (x)/(254)`
`:.` fraction of Zn remained unreacted
`=(((x)/(65)-(x)/(254)))/((x)/(65)) = 0.74`