Calculate Mass

AlohaIkepela · Post by **AlohaIkepela** » Mon Apr 18, 2016 8:24 pm

I'm so confused with this question: Calculate the mass of Fe₂O₃ needed to completely react with 1.64 x 10^12 g Mg₃(PO₄)₂

engineer7 · Post by **engineer7** » Wed May 18, 2016 4:45 am

Fe₂O₃ + Mg₃(PO₄)₂ = 2Fe(PO₄) + 3MgO

1 mol of Fe₂O₃ reacts with 1 mol of Mg(PO₄)₂.

Mg₃(PO₄)₂ - 262.86 g/mol
1.64 x 10^12g of Mg₃(PO₄)₂ equals to 6.24 x 10^9 moles,
same amount of moles of Fe₂O₃ needed - 6.24 x 10^9.
Fe₂O₃ molar mass is 159.69 g/mol,
so 6.24 x 10^9 moles of Fe₂O₃ weight 9.96 x 10^11 g.