Question

9) for the reaction
[ 2Al + 3H2SO4 → Al2(SO4)3 + 3 H2SO4] 6.0 g of Al (MW=26.98 g/mol) are added to 12.0 mL of 0.20 M of H2SO4

A) what is the limiting reagent in the above reaction?

b) how many miles of H2SO4 will be formed?

c) how many grams of Al2(SO4)3 (MW=342.15 g/mol) are produced in the reaction?

Answer 1

2Al + 3H2SO4 → Al2(SO4)3 + 3 H2

no of moles of Al = W/G.A.Wt

                            = 6/26.98   = 0.2224 moles

no of moles of H2SO4 = molarity * volume in L

                                      = 0.2*0.012 = 0.0024 moles

2 moles of Al react with 3 moles of H2SO4

0.2224 moles of Al react with = 3*0.2224/2 = 0.3336 moles of H2SO4

H2SO4 is limiting reactant

3 moles of H2SO4 react with Al to gives 1 mole of Al2(SO4)3

0.3336 moles of H2So4 react with Al to gives = 1*0.3336/3 = 0.1112 moles of Al2(SO4)3

mass of Al2(So4)3 = no of moles* gram molar mass

                                = 0.1112*342.15   = 38.05 g >>>>answer

3 moles of H2SO4 react with Al to gives 3 moles of H2

0.3336 moles of H2SO4 react with Al to gives 0.3336 moles of H2

0.3336 moles of H2