Question

Calculate the number of moles 3.16 g of KMnO4 32.33 g of K2CrO4 100 g of KHCO3

Answer 1

The number of moles:
3.16g of
`KMnO_(4)`
1 mole of
`KMnO_(4)`- 158g.
x moles of
`KMnO_(4)`- 3.16g.

`x=(1\ mole*3.16g)/(158g)=(3.16)/(158)=0.02\ mole`

The number of moles:
32.33g of
`K_(2)CrO_(4)`
1 mole of
`K_(2)CrO_(4)`- 194g.
x moles of
`K_(2)CrO_(4)`- 32.33g.

`x=(1\ mole*32.33g)/(194g)=(32.33)/(194)=0.17\ mole`

The number of moles:
100g of
`KHCO_(3)`
1 mole of
`KHCO_(3)`- 100.1g.
x moles of
`KHCO_(3)`- 100g.

`x=(1\ mole*100g)/(100.1g)=(100)/(100.1)=0.999\ mole`


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