Question

49g of KClO3 is heated strongly.Calculate the amount of the solid and the gaseous product formed during the reaction.(K=39,Cl=35.5,O=16)​

Answer 1

Answer : The amount of the solid $(KCl)$ and the gaseous $(O_2)$ product formed during the reaction are, 29.8 grams and 19.2 grams respectively.

Explanation : Given,

Mass of $KClO_3$ = 49 g

Molar mass of $KClO_3$ = 122.5 g/mole

Molar mass of KCl = 74.5 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $KClO_3$.

$\text{Moles of }KClO_3=\frac{\text{Mass of }KClO_3}{\text{Molar mass of }KClO_3}=\frac{49g}{122.5g/mole}=0.4mole$

Now we have to calculate the moles of $(KCl)$ and $(O_2)$.

The balanced chemical reaction will be,

$2KClO_3\rightarrow 2KCl+3O_2$

From the balanced reaction we conclude that,

As, 2 mole of $KClO_3$ react to give 2 mole of $KCl$

So, 0.4 mole of $KClO_3$ react to give 0.4 mole of $KCl$

And,

As, 2 mole of $KClO_3$ react to give 3 mole of $O_2$

So, 0.4 mole of $KClO_3$ react to give $\frac{3}{2}\times 0.4=0.6mole$ of $O_2$

Now we have to calculate the mass of $(KCl)$ and $(O_2)$.

$\text{Mass of }KCl=\text{Moles of }KCl\times \text{Molar mass of }KCl$

$\text{Mass of }KCl=(0.4mole)\times (74.5g/mole)=29.8g$

And,

$\text{Mass of }O_2=\text{Moles of }O_2\times \text{Molar mass of }O_2$

$\text{Mass of }O_2=(0.6mole)\times (32g/mole)=19.2g$

Therefore, the amount of the solid $(KCl)$ and the gaseous $(O_2)$ product formed during the reaction are, 29.8 grams and 19.2 grams respectively.