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RESPECTED SIR/MADAM,PLEASE HELP ME OUT. A mixture of NaI and NaCl when heated with...
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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)`
`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`
Posted by llltkl on June 4, 2013 at 4:47 PM (Answer #1)
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