Question

If 50g copper oxide is added to 0.5mol sulphuric acid, calculate the unreacted copper oxide​

Answer 1

Answer: 10.3 grams of Copper oxide is left unreacted.

Step-by-step explanation:

To calculate the moles :

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

`\text{Moles of} CuO=(50g)/(79.5g)=0.63moles`

The balanced chemical reaction is:

`CuO+H_2SO_4\rightarrow CuSO_4+H_2O`

According to stoichiometry :

1 moles of
`H_2SO_4` require = 1 mole of
`CuO`

Thus 0.5 moles of
`H_2SO_4` will require=
`(1)/(1)* 0.5=0.5moles` of
`CuO`

Thus
`H_2SO_4` is the limiting reagent as it limits the formation of product and
`CuO` is the excess reagent.

moles of CuO left = (0.63-0.5) mol= 0.13 mol

Mass of
`CuO` left =
`moles* {\text {Molar mass}}=0.13moles* 79.5g/mol=10.3g`

10.3 grams of Copper oxide is left unreacted.