Question

8.If 3.20 g of Fe(OH)2 is treated with 2.50 g of phosphoric acid, what is the limiting reagent and what is the reactant in excess?                                           /7

9.How many grams of Fe3(PO4)2 precipitate can be formed?                       /2

10.If 3.99 g of Fe3(PO4)2 is actually obtained, what is the percent yield?               /2

Answer 1

$Moles (Fe(OH)_2)=\frac{3.20}{89.86}= 0.0356$

$Moles (H_3PO_4)=\frac{2.50}{97.994}=0.0255$

Hence Fe(OH)₂ is the excess reactant and phosphoric acid is the limiting reactant

3Fe(OH)₂ + 2H₃PO₄ → Fe₃(PO₄)₂ + 6H₂O

Moles of Fe₃(PO₄)₂$=\frac{0.0356}{3}=0.012moles$ Mass of Fe₃(PO₄)₂$=0.012×357=4.284 g$