Asked by Bayisaa
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.
Answers
Answered by
DrBob222
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 = ?
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 = ?
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