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

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Answer 1 · 2018-07-04T14:20:02

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

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"#