Question

For each of the following reactions, 27.0 g of each reactant is present initially.

Part A

Determine the limiting reactant.
2Al(s)+3Br2(g)→2AlBr3(s)

Part B

Calculate the grams of product in parentheses that would be produced.
(AlBr3)

Part C

Determine the limiting reactant.
4NH3(g)+5O2(g)→4NO(g)+6H2O(g)

Part D

Calculate the grams of product in parentheses that would be produced.
(NO)

Part E

Determine the limiting reactant.
CS2(g)+3O2(g)→CO2(g)+2SO2(g)

Part F

Calculate the grams of product in parentheses that would be produced.
(SO2)

Answer 1

no of moles of Al   = W/G.A.Wt

                             = 27/27 = 1 moles

no of moles of Br2 = W/G.M.Wt

                             = 27/160   = 0.16875moles

2Al(s)+3Br2(g)→2AlBr3(s)

2 moles of Al react with 3 moles of Br2

1 mole of Al react with = 3*1/2 = 1.5 moles of Br2 is required

Br2 is limiting reactant

Part-B

2Al(s)+3Br2(g)→2AlBr3(s)

3 moles of Br2 react with excess of Al to gives 2 moles of AlBr3

0.16875 moles of Br2 react with excess of Al to gives = 2*0.16875/3   = 0.1125 moles of AlBr3

mass of AlBr3   = no of moles * gram molar mass

                        = 0.1125*266.69 = 30g of AlBr3 >>>>answer

part-C

no of moles of NH3 = W/G.M.Wt

                               = 27/17   = 1.588 moles

no of moles of O2   = W/G.M.Wt

                              = 27/32   = 0.84375

4NH3(g)+5O2(g)→4NO(g)+6H2O(g)

4 moles of NH3 react with 5 moles of O2

1.588 moles of NH3 react with = 5*1.588/4 = 1.985 moles of O2 is required

O2 is limiting reactant

part-D

4NH3(g)+5O2(g)→4NO(g)+6H2O(g)

5 moles of O2 react with excess of NH3 to gives 4 moles of NO

0.84375 moles of O2 react with excess of Nh3 to gives = 4*0.84375/5   = 0.675 moles of NO

mass of NO = no of moles * gram molar mass

                     = 0.675*30 = 20.25g of NO

part-E

no of moles of CS2   = W/G.M.Wt

                                 = 27/76   = 0.3553 moles

no of moles of O2   = W/G.M.Wt

                                = 27/32   = 0.84375moles

CS2(g)+3O2(g)→CO2(g)+2SO2(g)

1 moles of CS2 react with 3 moles of O2

0.3553 moles of CS2 react with = 3*0.3553/1   = 1.0659moles of O2 is required

O2 is limiting reactnat

part-F

CS2(g)+3O2(g)→CO2(g)+2SO2(g)

3 moles of O2 react with excess of CS2 to gives 2 moles of SO2

0.84375 moles of O2 react with excess of CS2 to gives = 2*0.84375/3   = 0.5625 moles of SO2

mass of SO2 = no of moles * gram molar mass

                      = 0.5625*64   = 36g >>>>answer