Question

4.9 gm of potassium chlorate Heated strongly calculate the amount of solid products form and gaseous products​

Answer 1

Answer : The amount of KCl and $O_2$ are, 2.925 g and 1.888g respectively.

Explanation : Given,

Mass of $KClO_3$ = 4.9 g

Molar mass of $KClO_3$ = 123 g/mole

Molar mass of $KCl$ = 75 g/mole

Molar mass of $O_2$ = 32 g/mole

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

$\text{Moles of }KClO_3=\frac{\text{Mass of }KClO_3}{\text{Molar mass of }KClO_3}=\frac{4.9g}{123g/mole}=0.039mole$

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

The balanced chemical reaction will be,

$2KClO_3(s)\rightarrow 2KCl(s)+3O_2(g)$

From the balanced chemical reaction, we conclude that

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

So, 0.039 moles of $KClO_3$ react to give 0.039 moles of KCl

The moles of KCl = 0.039 mole

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

So, 0.039 moles of $KClO_3$ react to give $\frac{3}{2}\times 0.039=0.059$ moles of $O_2$

The moles of $O_2$ = 0.059 mole

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

$\text{Mass of }KCl=\text{Moles of }KCl\times \text{Molar mass of }KCl=0.039mole\times 75g/mole=2.925g$

$\text{Mass of }O_2=\text{Moles of }O_2\times \text{Molar mass of }O_2=0.059mole\times 32g/mole=1.888g$

Therefore, the amount of KCl and $O_2$ are, 2.925 g and 1.888g respectively.