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
"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
Comments
Leave a comment