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

Step-by-step 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 =(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 =
`(3)/(2)* 0.82=1.23` moles of
`O_2`

Volume of
`O_2=moles* {\text {molar volume}}=1.23* 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