Question

How many moles are in 198.34 g of Fe3(PO4)2?

Answer 1

Approximately `0.555` moles of iron(II) phosphate.

Explanation:

To find the number of moles, we use the formula:

`n=m/M`

where:

• `n` is the number of moles of substance

• `m` is the mass of the substance in the given sample

• `M` is the molar mass of the substance

For iron(II) phosphate `[Fe_3(PO_4)_2)]`, it has a molar mass of `357.48 \ "g/mol"`. So, we get:

`n=(198.34color(red)cancelcolor(black)"g")/(357.48color(red)cancelcolor(black)"g""/mol")`

`~~0.555 \ "mol"`