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Q. A mixture containing `KClO_3,KHCO_3,K_2CO_3` and `KCl` was heated, producing...

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Q. A mixture containing `KClO_3,KHCO_3,K_2CO_3` and `KCl` was heated, producing `CO_2,O_2 ` and `H_2O` gases according to the following equations:-

   `2KClO_3 (s) -> 2KCl(s) + 3O_2`

   `2KHCO_3(s) -> K_2O(s) +H_2O(g) +2CO_2(g)`

   `K_2CO_3(s) -> K_2O(s) +CO_2(g)`

The `KCl` doesn't react under the conditions of the reaction. If 100.0 g of the mixture produces 1.80 g of `H_2O,` 13.20 g of `CO_2` and 4.0 g of `O_2` ,what was the composition of the original mixture?

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Let the mixture contains x g `KClO_3` , y g `KHCO_3` , and z g `K_2CO_3` . Mass of KCl was, thus [100-(x+y+z)] g.

Molar mass of the relevant species are:







According to stoichiometry of the given decomposition reactions,

Amount (in g) of `H_2O` formed

`=18/200.2 y`

Amount (in g) of `CO_2` formed

`=(2*44)/200.2 y+44/138.2 z`

Amount (in g) of `O_2` formed

`=(3*32)/245.1 x`

By the given conditions

`18/200.2y=1.8`   --- (i)

`rArr y=(1.8*200.2)/18=20.02`

Again, `(3*32)/245.1 x=4.0` --- (ii)

`rArr x=(4*245.1)/(3*32)=10.21`

And, `(2*44)/200.2 y+44/138.2 z=13.2` --- (iii)

Plugging in the value of y,

`(2*44)/200.2*20.02+44/138.2 z=13.2`

`rArr 44/138.2z=13.2-(2*44*20.02/200.2)=4.4`

`rArr z=(4.4*138.2)/44=13.82`

As the total mixture was 100 g, these values represent corresponding percentage compositions.

Thus, the percentage composition of the given mixture is 10.21 percent `KClO_3` , 20.02 percent `KHCO_3` , 13.82 percent `K_2CO_3` , and 55.95 percent KCl.

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Posted (Reply #1)

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Awesome answer.Thanks a lot.

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