5.6g of iron metal Fe are heated strong and mixed with 8g of sulfur S to form 4.4g of iron(ll) sulfide FeS. a, identify the limiting and excess reactants. b, calculate the mass of FeS formed. C,calculet the percent yield of FeS.

Answer 1

Fe + S --> FeS
a. moles Fe = g/atomic mass = 5.6/56 = 0.1 mol
moles S = 8/32 = 0.25
0.1 mole Fe will require 0.1 mol S and you have that much; therefore, Fe is the limiting regent (LR) and S is the excess reagent (ER).

b. I think you meant how much FeS COULD BE FORMED since you know from the problem that 4.4 g was formed; i.e., ou want to know the theoretical yield. 0.1 mol Fe will produce 0.1 mol FeS. grams FeS produced = mols x molar mass = 0.1 mole x 88 = 8.8 grams FeS. 8.8 g FeS is the theoretical yield.

c. % yield = (actual yield/theoretical yield) *100 = ?