Question

Molecular oxygen is produced by the decomposition of KClO3. How many liters of O2 are produced by the complete decomposition of 100.0 g of KClO3 at STP?

Answer 1

Answer: 27.5 liters

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$  

For $KClO_3$

Given mass = 100.0 g

Molar mass of  $KClO_3$ = 122.5 g/mol

Putting values in above equation, we get:

$\text{Moles of}KClO_3 =\frac{100}{122.5}=0.82moles$

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

2 moles of $KClO_3$ produces 3 moles of $O_2$

0.82 moles of $KClO_3$ produces =$\frac{3}{2}\times 0.82=1.23$ moles of $O_2$

Volume of $O_2=moles\times {\text {molar volume}}=1.23\times 22.4=27.5L$

Thus 27.5 liters of $O_2$ are produced by the complete decomposition of 100.0 g of $KClO_3$ at STP