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

Question

5267 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-04T14:20:02

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

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

$n = \frac{m}{M}$ $n=m/M$

where:

$n$ $n$ is the number of moles of substance

$m$ $m$ is the mass of the substance in the given sample

$M$ $M$ is the molar mass of the substance

For iron(II) phosphate $[F e_{3} {(P O_{4})}_{2})]$ $[Fe_3(PO_4)_2)]$ , it has a molar mass of $357.48 \ "g/mol"$ . So, we get:

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

$~~0.555 \ "mol"$