Question

2KMnO4= K2MnO4+ MnO2+O2 how many grams of KMnO4 are required to produce 1.60 grams of O2

Answer 1

Answer: 15.8 g of
`KMnO_4` will be required to produce 1.60 grams of
`O_2`

Step-by-step explanation:

To calculate the moles :

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

`\text{Moles of} O_2=(1.60g)/(32g/mol)=0.05moles`

`2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2`

According to stoichiometry :

As 1 mole of
`O_2` is given by = 2 moles of
`KMnO_4`

Thus 0.05 moles of
`O_2` is given by =
`(2)/(1)* 0.05=0.10moles` of
`KMnO_4`

Mass of
`KMnO_4=moles* {\text {Molar mass}}=0.10moles* 158g/mol=15.8g`

Thus 15.8 g of
`KMnO_4` will be required to produce 1.60 grams of
`O_2`