Question

How do you calculate percent yield? Step By Step.
I'll give an equation.

Answer 1

In this reaction, 200.0 g of FeS2 is burned in 100.0 g of oxygen, and 55.00 g of Fe2O3 is produced.

4FeS2 + 11O2 -> Fe2O3 + SO2

What is the percent yield of Fe2O3?

Answer 2

percent yield=practical amount/theoretical amount

Answer 3

first determine limiting reactant then use stoichiometric coefficients to obtain Fe2O3 amount. mass of Fe2O3 that mentioned in question is practical amount and mass of Fe2O3 that you calculate is theoretical amount. got it?

Answer 4

did you understand what i said?

Answer 5

No.

Answer 6

do you know how to determine limiting reactant?

Answer 7

First, the equation must be balanced:
4 FeS2 + 11 O2 = 2 Fe2O3 + 8 SO2
Convert to moles:
(200.0 g FeS2) / (119.9750 g FeS2/mol) = 1.66701 mol FeS2
(100.0 g O2) / (31.99886 g O2/mol) = 3.12511 mol O2

3.12511 moles of O2 would react completely with 3.12511 x (4/11) = 1.13640 moles of FeS2, but there is more FeS2 present than that, so FeS2 is in excess and O2 is the limiting reactant.

(3.12511 mol O2) x (2 mol Fe2O3 / 11 mol O2) x (159.6882 g Fe2O3/mol) =
90.735 g Fe2O3 in theory

(55.00 g Fe2O3 actual) / (90.735 g Fe2O3 in theory) = 0.6062 =
60.62% yield Fe2O3

Answer 8

I am so sorry, but this is where I'm lost in chemistry....what the crap did you just do and how????

Answer 9

I don't know how to make it plainer.
moradi312: Do you want to take over again?

Answer 10

Sorry @Tech1 I don't know why I just can't catch on. :(