Question

A reaction produces 92.50 g FeSO4. How many grams of CuSO4 are necessary for this to occur?

Fe+CuSO4⟶Cu+FeSO4

Answer 1

Answer : The mass of
`CuSO_4` needed will be, 97.2 grams.

Explanation : Given,

Mass of
`FeSO_4` = 92.50 g

Molar mass of
`FeSO_4` = 151.908 g/mole

Molar mass of
`CuSO_4` = 159.609 g/mole

First we have to calculate the moles of
`FeSO_4`.

`\text{Moles of }FeSO_4=\frac{\text{Mass of }FeSO_4}{\text{Molar mass of }FeSO_4}=(92.50g)/(151.908g/mole)=0.609moles`

Now we have to calculate the moles of
`CuSO_4`.

The balanced chemical reaction is,

`Fe+CuSO_4\rightarrow Cu+FeSO_4`

From the balanced reaction we conclude that

As, 1 mole of
`FeSO_4` obtained from 1 mole of
`CuSO_4`

So, 0.609 moles of
`FeSO_4` obtained 0.609 moles of
`CuSO_4`

Now we have to calculate the mass of
`CuSO_4`.

`\text{Mass of }CuSO_4=\text{Moles of }CuSO_4* \text{Molar mass of }CuSO_4`

`\text{Mass of }CuSO_4=(0.609mole)* (159.609g/mole)=97.2g`

Therefore, the mass of
`CuSO_4` needed will be, 97.2 grams.

Answer 2

Fe+CuSO4⟶Cu+FeSO4

Given that

FeSO4 = 92.50 g

Number of moles = amount in g / molar mass

=92.50 g / 151.908 g/mol

=0.609 moles FeSO4

Now calculate the moles of CuSO4 as follows:

0.609 moles FeSO4 * 1 mole CuSO4 /1 mole FeSO4

= 0.609 moles CuSO4

Amount in g = number of moles * molar mass

= 0.609 moles CuSO4 * 159.609 g/mol

= 97.19 g CuSO4