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)

Explanation

Let x g be the initial weight of the Zn metal and iodine each. Since `I_(2)` is completely converted to `ZnL_(2)`, we have,
initial no. of moles : `{:((x)/(65)" "(x)/(254)" "0),(Zn " "+ " "I_(2)" "to " "ZnI_(2)):}`
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.`