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?
1 Answer | Add Yours
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
Amount (in g) of `CO_2` formed
`=(2*44)/200.2 y+44/138.2 z`
Amount (in g) of `O_2` formed
By the given conditions
`18/200.2y=1.8` --- (i)
Again, `(3*32)/245.1 x=4.0` --- (ii)
And, `(2*44)/200.2 y+44/138.2 z=13.2` --- (iii)
Plugging in the value of y,
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.
Awesome answer.Thanks a lot.
We’ve answered 330,800 questions. We can answer yours, too.Ask a question