Question

aluminum nitrate reacts with calcium sulfate to produce aluminum sulfate and calcium nitrate. If 4.4 grams of aluminum nitrate and 1.9 grams of calcium sulfate are mixed and allowed to react to completion, how many grams of calcium nitrate could be produced?

Answer 1

2Al(NO3)3 + 3CaSO4 -------------> 3Ca(NO3)2 + Al2(SO4)3

no of moles of Al(NO3)3   = W/G.M.Wt

                                         = 4.4/213   = 0.0206 moles

no of moles of CaSO4   = W/G.M.Wt

                                      = 1.9/136 = 0.0139moles

2 moles of Al(NO3)3 react with 3 moles of CaSO4

0.209 moles of Al(NO3)3 react with = 3*0.209/2   = 0.3135 moles of CaSo4 is required

CaSO4 is limiting reactant

3 moles of CaSo4 react with Al(NO3)3 to gives 3 moles of Ca(NO3)2

0.0139 moles of CaSO4 react with Al(NO3)3 to gives = 3*0.0139/3   = 0.0139 moles of Ca(NO3)2

mass of Ca(NO3)2 = no of moles * gram molar

                                = 0.0139*164   = 2.28g of Ca(NO302 >>>>>answer