Question

$4.90$ grams of $KCl{O_3}$ on heating, shows a weight loss of $0.384$ grams. What percent of the original $KCl{O_3}$ has decomposed?

Answer 1

Hint : In the above given question we have sodium chlorate which when heated decomposes to potassium chloride and oxygen gas. The gas oxygen decomposes and that must be the reason for the mass loss after heating of potassium chlorate.

Complete Step By Step Answer:
In the above given question, we have been asked what is the mass loss in potassium chlorate after it is heated.
Now first we need to analyse the reaction of potassium chlorate in which it gets decomposed.
$2KCl{O_3}\xrightarrow{\Delta }2KCl + 3{O_2}$
Now in the above reaction we see that potassium chlorate decomposes into potassium chloride and oxygen gas on heating. Now this oxygen is a gas which gets blown away in the form of gas which results in some mass loss.
So, we know that the mass loss is equal to the moles of oxygen formed.
Mass loss is equal to $0.384$ grams, which is the mass of oxygen formed which when divided by the molecular weight of the oxygen gas would get the moles of oxygen formed.
Moles of oxygen: -
$moles = \dfrac{{given\,mass}}{{molecular\,mass}}$
$\Rightarrow moles = \dfrac{{0.384}}{{32}}$
$\Rightarrow moles = 0.012$
Now from the reaction $2KCl{O_3}\xrightarrow{\Delta }2KCl + 3{O_2}$ ,we know that two moles of potassium chlorate forms three moles of oxygen gas, now we know that the ratio of the formation of moles of potassium chlorate and oxygen gas is going to be constant, therefore using the ratio and the calculates moles of oxygen, we get that
$moles\,of\,KCl{O_3} = \dfrac{2}{3} \times moles\,of\,{O_2}$
$\Rightarrow moles\,of\,KCl{O_3} = \dfrac{2}{3} \times 0.012$
$\Rightarrow moles\,of\,KCl{O_3} = 0.008$
Now the moles of potassium chlorate we have, on multiplying the moles of potassium chlorate decomposed by the molar mass of potassium chlorate we get
$KCl{O_3}\,decomposed = molar\,mass\,of\,KCl{O_3} \times moles\,of\,KCl{O_3}\,decomposed$
$\Rightarrow KCl{O_3}\,decomposed = 122.5 \times 0.008$
$\Rightarrow \,KCl{O_3}\,decomposed = 0.9804\,grams$
Now we need to calculate the percentage decomposed out of the whole sample of $4.90$ grams of $KCl{O_3}$
$\% \,of\,original\,sample = \dfrac{{0.98}}{{4.90}} \times 100$
$\Rightarrow \% \,of\,original\,sample = 20\%$
$20\%$ is the potassium chlorate decomposed in the reaction when $4.90$ grams of $KCl{O_3}$ is heated and there is a weight loss of $0.384$ grams.

Note :
Potassium chlorate is widely used for the making of very different products. One of the products it is used to form are noise makers in crackers on mixing it with silver fulminate. It is also used as an oxidizer in smoke grenades and creates oxygen gas which gives a pop sound.