RESPECTED SIR/MADAM,PLEASE HELP ME OUT.

A mixture of NaI and NaCl when heated with H2SO4 produced the same weight of sodium sulphate as that of the original mixture.Calculate the percentage of NaI in the mixture. (Ans: 28.85%)

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Both NaI and NaCl upon treatment with H2SO4 produce Na2SO4 according to the following equations:

2NaI + H2SO4 → Na2SO4 + 2HI --- (i)

2NaCl + H2SO4 → Na2SO4 + 2HCl --- (ii)

Let ‘x’ g NaI was taken initially in a total mixture of ‘y’ g (NaI+NaCl). Therefore, mass of NaCl initially taken was (y-x) g.

By stoichiometry of equation (i), upon treatment with H2SO4,

`2*(23+127)= 300` g NaI produces `(2*23+32+4*16)= 142` g Na2SO4.

Or, 'x' g NaI produces `(142*x)/300 = 0.473333*x` g Na2SO4

Similarly, by stoichiometry of equation (ii), upon treatment with H2SO4,

`2*(23+35.5)= 117` g NaCl produces `(2*23+32+4*16)= 142` g Na2SO4.

Or, `(y-x)` g NaCl produces `(142*x)/117 = 1.213675*(y-x)` g Na2SO4

By condition of the problem,

`0.473333*x + 1.213675*(y-x) = y`

`rArr` `(x-0.473333*x) = (1.213675*y-y)`

`rArr`

`0.740342*x = 0.213675*y `

`rArr` `x/y = 0.213675/0.740342 = 0.2886`

**Therefore percentage of NaI in the mixture was `(x*100)/y` = `0.2886*100` = `28.86` **

**Sources:**

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