2Al(s) + 3F2 ------------> 2AlF3 no of moles of Al = W/G.A.Wt = 1.75/27 = 0.0648

Question

2Al(s) + 3F2 ------------> 2AlF3

no of moles of Al     = W/G.A.Wt

                                = 1.75/27   = 0.0648 moles

no of moles of F2   = W/G.M.Wt

                              = 3.23/38   = 0.085 moles

2 moles of Al react with 3 moles of F2

0.085 moles of Al react with = 3*0.085/2   = 0.1275moles of F2

F2 is limiting reactant

3 moles of F2 react with Al to gives 2 moles of AlF3

0.085 moles of F2 react with Al to gives = 2*0.085/3 = 0.056 moles of AlF3

mass of AlF3 = no of moles * gram molar mass

                      = 0.056*84   = 4.704g of AlF3

mass of AlF4 = 4.704g >>>>>answer

Explanation / Answer

What mass of AIF3 will be produced from 1.75 g of Al and 3.23 g of F2?

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