Question

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

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 $FeSO_4$.

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

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 $CuSO_4$.

$\text{Mass of }CuSO_4=\text{Moles of }CuSO_4\times \text{Molar mass of }CuSO_4$

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

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