Question

How manymoles of each ion are present in 175 mL of 0.147 M Fe2(SO4)3?

Answer 1

Answer: 0.0257 moles of
`Fe^(3+)` and 0.0257 moles of
`SO_4^(2-)`

Step-by-step explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

`Molarity=\frac{moles}{\text {Volume in L}}`

moles of
`Fe_2(SO_4)_3=Molarity* {\text {Volume in L}}=0.147* 0.175L=0.0257moles`

The balanced reaction for dissociation will be:

`Fe_2(SO_4)_3\rightarrow Fe^(3+)+SO_4^(2-)`

According to stoichiometry:

1 mole of
`Fe_2(SO_4)_3` gives 1 mole of
`Fe^(3+)` and 1 mole of
`SO_4^(2-)`

Thus there will be 0.0257 moles of
`Fe^(3+)` and 0.0257 moles of
`SO_4^(2-)`