Question

If the decomposition of a sample of KClO3 produces 3.29 g of O2 , what was the mass (g) of the original sample?

Answer 1

Answer: Thus 8.41 g was the mass of the original sample

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} O_2=\frac{3.29g}{32g/mol}=0.103moles$

$2KClO_3\rightarrow 2KCl+3O_2$  

According to stoichiometry :

As 3 moles of $O_2$ are produced by = 2 moles of $KClO_3$

Thus 0.103 moles of $O_2$ are produced by =$\frac{2}{3}\times 0.103=0.0687moles$  of $KClO_3$

Mass of $KClO_3=moles\times {\text {Molar mass}}=0.0687moles\times 122.5g/mol=8.41g$

Thus 8.41 g was the mass of the original sample.