Q. A mixture of `NaCl` and `NaI` when heated with `H_2SO_4` , produced the same weight of sodium sulphate as that of the original mixture.Calculate percentage of `NaI` in the mixture.

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Let, the total amount of mixture of NaCl and NaI taken initially was x g, out of which amount of NaCl was y g. So, amount of NaI in the initial mixture was (x-y) g.

Balanced chemical reactions of NaCl and NaI with H2SO4 are as follows:

2NaCl + H2SO4 -> Na2SO4 + HCl

2NaI + H2SO4 -> Na2SO4 + HI

Amount of components obtainable/reacting freom stoichiometric considerations are as follows:

NaCl Na2SO4 NaI Na2SO4

------ --------- ---------- ------------

117 142 300 142

`y` `(142*y)/117` `(x-y) ` `(142*(x-y))/300`

By condition, `(142*y)/117+(142*(x-y))/300 = x` ` `

`rArr 14200y+5538x-5538y=11700x ` (LCM of 117 and 300 is 11700)

`rArr 8662y=6162x`

`(y/x)*100=(6162/8662)*100=71.14`

This is the percent (by mass) of NaCl present in the original mixture.

Therefore, percent (by mass) of NaI present in the original mixture was 100-71.14=28.86.

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