Question

A mixture containing $KCl{O_3},KHC{O_3},{K_2}C{O_3}$ and $KCl$ was heated, producing $C{O_2},{O_2}$ and ${H_2}O$ gases according to the following equations:
$\ 2KCl{O_{3(s)}} \to 2KC{l_{(s)}} + 3{O_2} \\\ 2KHC{O_{3(s)}} \to {K_2}{O_{(s)}} + {H_2}{O_{(g)}} + 2C{O_{2(g)}} \\\ {K_2}C{O_{3(s)}} \to {K_2}{O_{(s)}} + C{O_{2(g)}} \\\ \$
The $KCl$ does not react under the conditions of the reaction. If $100g$ of the mixture produces $1.80g$ of ${H_2}O$ , $13.20g$ of $C{O_2}$ and $4g$ of ${O_2}$ . Then what was the composition of the original mixture?

Answer 1

In order to find out the composition of the original mixture we have to calculate the number of moles of every element present in the reaction. After that compare them with the values as per given in the question. Doing so we will determine the mass of every element present in the mixture.

Complete step by step answer:
According to the question, we have three reactions that took place:
$\ 2KCl{O_{3(s)}} \to 2KC{l_{(s)}} + 3{O_2} \\\ 2KHC{O_{3(s)}} \to {K_2}{O_{(s)}} + {H_2}{O_{(g)}} + 2C{O_{2(g)}} \\\ {K_2}C{O_{3(s)}} \to {K_2}{O_{(s)}} + C{O_{2(g)}} \\\ \$
Let’s consider the reaction: $2KHC{O_3} \to {K_2}O + {H_2}O + 2C{O_2}$
As from all the reactions we can see that water molecules are producing only in this reaction.
So mass of total ${H_2}O$ is $1.80g$
So, we can write:
Weight of water produced = $\dfrac{{1.8}}{{18}} = 0.1mol$
Now, the number of moles of $KHC{O_3}$ = $0.1 \times 2 \Rightarrow 0.2mol$
So, weight of$KHC{O_3}$ = $0.2 \times 100g \Rightarrow 20g$
In the same equation number of moles of ${O_2}$ produced = $\dfrac{4}{{32}} \Rightarrow \dfrac{1}{8}mol$
Now, consider the equation: $2KCl{O_3} \to 2KCl + 3{O_2}$
It is clear that two mole of $KCl{O_3}$ produce three moles of oxygen molecules.
So we can write: $3 \times KCl{O_3}moles = 2 \times {O_2}moles$
So number of moles of $KCl{O_3}$ = $\dfrac{2}{3} \times$ number of moles of ${O_2}$
Number of moles of $KCl{O_3}$ = $\dfrac{2}{3} \times \dfrac{1}{8} \Rightarrow \dfrac{1}{{12}}mol$
Weight of $KCl{O_3}$ = $\dfrac{1}{{12}}(39 + 35.5 + 3 \times 16)g \Rightarrow \dfrac{1}{{12}} \times 122.5g \Rightarrow 10.2g$
And also number of moles of carbon dioxide produced = $\dfrac{{13.2}}{{44}}mol$
So, total number of moles of carbon dioxide = Mole of $KHC{O_3}$ + Mole of ${K_2}C{O_3}$
$\Rightarrow \dfrac{{13.2}}{{44}} = 0.2 + moles{K_2}C{O_3}$
So moles of ${K_2}C{O_3}$ is: $\dfrac{{13.2}}{{44}} - 0.2$
$\Rightarrow 0.1mol$
So, weight of ${K_2}C{O_3}$ = $0.1 \times 138g \Rightarrow 13.8g$

Note:
There is a simple relation between the number of moles and weight of compounds. Number of moles is the ratio of given weight of compound and molecular mass of compound. On applying this we will get the number of moles of compound present in the given weight of compound. Avogadro states that one mole of any compound consists of a definite number of constituents that is $6.02 \times {10^{23}}$ number of constituents. These constituents may be atoms, ions or molecules. This definite value is called Avogadro constant.