Question

How many fe(ii) ions are there in 15.0 g of feso4?

Answer 1

5.94 x 10^22 iron(II) ions

Answer 2

Answer : The number of Fe(II) ions present are
`5.94* 10^(22)`

Explanation : Given,

Mass of
`FeSO_4` = 15.0 g

Molar mass of
`FeSO_4` = 152 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}=(15.0g)/(152g/mole)=0.0987mole`

Now we have to calculate the number of Fe(II) ions.

In
`FeSO_4`, there 1 atom of iron ion and 1 atom of sulfate ion.

As we know that,

1 mole of substance always contains
`6.022* 10^(23)` number of atoms or ions.

As, 1 mole of
`FeSO_4` contains
`6.022* 10^(23)` number of Fe(II) ions.

So, 0.0987 mole of
`FeSO_4` contains
`0.0987* 6.022* 10^(23)=5.94* 10^(22)` number of Fe(II) ions.

Therefore, the number of Fe(II) ions present are
`5.94* 10^(22)`