Question

If 12.6 grams of iron (III) oxide reacts with 9.65 grams of carbon monoxide to produce 7.23 g of pure iron, what are the theoretical yield and percent yield of this reaction? Be sure to show the work that you did to solve this problem. unbalanced equation: Fe2O3 + CO “yields”/ Fe + CO2

Answer 1

OMG it's chemistry!!!

Answer 2

First write a balanced equation

Answer 3

the balanced eq is Fe2O3 +3CO ==> 3CO2 + 2Fe

Answer 4

And then find your limiting and excess reactants

Answer 5

\[Fe _{2}O _{3}+3CO \rightarrow2 Fe+3CO _{2}\]

Answer 6

i can't , can u please help me out with this?

Answer 7

Find the moles of ferric oxide

Answer 8

It is 159.6882

Answer 9

That's the molecular weight. Find the number of moles

Answer 10

okay i get the whole procedure now, thank u for the help :)

Answer 11

Divide 12.6 by the molecular weight to find the number of moles of ferric oxide.

Answer 12

Then divide 9.65 by 28.01 to find the moles of CO

Answer 13

Then you will find that the ferric oxide is the limiting reactant.

Answer 14

number of moles for ferric acid is 0.07890

Answer 15

yes

Answer 16

moles of CO are 0.34451

Answer 17

Yes.

Answer 18

According to the balanced equation 1 mole of ferric oxide reacts completely with 3 moles of CO. We have 0.07890 moles of ferric oxide. That would require 3 times .07890 moles of CO which is .2367 moles

Answer 19

So we have more than enough CO since we have .34451 moles

Answer 20

So all of the ferric oxide will be used up in the reaction.

Answer 21

And since 1 mole of ferric oxide produces 2 moles of pure iron, according to the balanced equation, we will get 2 x .07890 moles of pure iron.

Answer 22

That is .1578 moles of pure iron.
One mole of pure iron is 55.845 so .1578 moles is .1578 x 55.845 or 8.8123 and that is the theoretical yield.

Answer 23

Since the actual yield was 7.23 the percent yield is 7.23/8.8123 or .82044 or 82.04 %

Answer 24

oh now i get how it works.... btw thank you very much ! :)

Answer 25

yw