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

Jul 4, 2018

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

#### Explanation:

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

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

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 $357.48 \setminus \text{g/mol}$. So, we get:

$n = \left(198.34 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g")/(357.48color(red)cancelcolor(black)"g""/mol}}}}\right)$

$\approx 0.555 \setminus \text{mol}$