Question

if 7.65 grams of iron (iii) oxide react with 6.85 grams of carbon monoxide to produce 5.10g of pure iron, what are the theoretical yield and percent of this reaction?

Answer 1

Theoretical yield is the amount of iron produced at 100% yield (meaning everything reacted together completely) and the actual yield of iron in the question is 5.10g. So, we need to find the theoretical yield, and then work out the percentage yield when 5.10g of iron is produced.

So firstly you need to write a balanced equation for the reaction:
Fe2O3 + 3 CO = 3 CO2 + 2 Fe

Now, you need to work out the mass of Fe, however we dont know that at 100% yield, so we will have to work out the moles of iron (iii) oxide or carbon monoxide, and then use stoichiometry to find the moles of Fe. For this example we will use carbon monoxide.

So, to find the moles of CO (carbon monoxide) you will have to do: mass(6.85g)/molar mass(on periodic table, add mass number of carbon and oxygen together) = moles

Once you have the moles of CO, now you can find the moles of Fe. Now, CO has a '3' in front of it, and Fe has a 2. The amount of moles we got for CO is for THREE lots of carbon monoxide, and we have TWO lots of iron.
So we will need to take our moles for CO and divide it by 3 to get ONE lot of moles (CO moles/3), and then times it by 2 to get the moles for TWO lots of Fe. Then you will have the moles of Iron.

Now, you will need to find the mass of iron, and you can do that by doing: moles(of Fe) x molar mass(atomic number of Fe on periodic table) = mass (g)
Then you will have your theoretical yield in grams :)

Now, to work out the percentage yield, you will have to use this equation:
\[\left( \frac{ Actual yield }{ Theoretical yield } \right) \times 100\]

Actual yield being the yield of mass of Fe in the question (5.10g)/
Theoretical yield being what you just worked out
x 100
= (your answer)%

And now you should have your percentage yield. I hope this helps!