1.

Equal weights of Zn metal and iodine are mixed together and the iodine is completelyconverted to ZnI_(2). What fraction by weightof the original zinc remains unreacted?(zn = 65 , I = 127 )

Answer»

Solution :Let x g be the INITIAL WEIGHT of the Znmetaland iodine each. SINCE `I_(2)` is completelyconverted 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.`


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