Question

2Fe2O3(s)+3 CO2(g)+4 Fe(s)

A) if 2.70 g of Fe2O and 4.05 of carbon react, what mass of iron is expected to form?

B) what was the limiting reactant?

C) How much excess reagent (in grams) is left at the end of the reaction?

D) if percent yield is 92%, what mass of iron was actually obtained in the reaction? (Actual yield?)

Answer 1

2Fe2O3(s)+3C----->3 CO2(g)+4 Fe(s)

no of moles of Fe2O3 = W/G.M.Wt

                                   = 2.7/160   = 0.0168 moles

no of moles of C        = 4.05/12 = 0.3375 moles

2 moles of Fe2O3 react with 3 moles of C

0.0168 moles of Fe2O3 react with = 3*0.0168/2   = 0.0252 moles of C

limiting reagent is Fe2O3

Excess reagent is C

the amount of excess reagent left = 0.3375 -0.0252 =0.3123 moles = 0.3123* gram molar mass = 0.3123*12 =3.7476gm

2 mole of Fe2O3 react with C to form 4 moles of Fe

0.0168moles of Fe2O3 react with C to form =4*0.0168/2 = 0.0336 moles of Fe

mass of Fe = no of moles of Fe*atomic weight of Fe

                 = 0.0336*56 =1.8816gm of Fe

Actual yield = percent yield*theretical yield

                   = 92*1.8816/100 = 1.731 gm >>>> answer