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)rarrZnI_(2)`
Initial no. of moles `(x)/(65)(x)/(254)" "0`
No. of moles of the end of the reaction :
`((x)/(65)-(x)/(254))" 0 "(x)/(254)`
`therefore " fraction of Zn remained unreacted"`
`=(((x)/(65)-(x)/(254)))/((x)/(65))=0.74`