Answer to Question #183484 in General Chemistry for Krissy

Question #183484

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


1
Expert's answer
2021-04-22T03:04:26-0400

"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\u00d7357=4.284 g"




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