Question

For the following reaction, 27.7 grams of diphosphorus pentoxide are allowed to react with 13.7 grams of water.

diphosphorus pentoxide (s) + water (l) ----> phosphoric acid (aq)

What is the maximum amount of phosphoric acid that can be formed?

grams

What is the FORMULA for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

P2O5 + 3H2O --------------> 2H3PO4

no of moles of P2O5    = W/G.M.Wt

                                     = 27.7/142   = 0.195 moles

no of moles of H2O    = W/G.M.Wt

                                   = 13.7/18 = 0.76 moles

from balanced equation

1 moles of P2O5 react with 3moles of H2O

0.195 moles of P2O5 react with = 3*0.195/1   = 0.585 moles of H2O is required

P2O5 is limiting reagent

1 mole of P2O5 react with H2O to gives 2 moles of H2PO4

0.195 moles of P2O5 react with H2O to gives = 2*0.195/1   = 0.39 moles H3PO4

mass of H3PO4 = no of moles * gram molar mass

                           = 0.39*98   = 38.22g of H3PO4