Question

If 7.65 grams of iron (III) oxide reacts with 6.85 grams of carbon monoxide to produce 5.10 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

So first balance the equation.

Answer 2

okay can you give me a second to balance it because this one is a little harder than the last one

Answer 3

take your time

Answer 4

Okay I think it is Fe203 + 3CO "yields" 2Fe + 3CO2

Answer 5

Its good! Now convert the masses given to moles.

We need to determine which is the limiting reactant before continuing.

Answer 6

okay just give a minute and again thanks for taking your time to help me

Answer 7

yeah dude, no problem. you dont have to tell me to wait a minute, i know it takes some time.

Answer 8

I know you start with the given Grams which is 7.65 but I forgot where to go from there

Answer 9

I think then you need to add the elements masses together?

Answer 10

divide the mass (m) by the molar mass (M) to get moles (n): \(n=\dfrac{m}{M}\)

Answer 11

I am having trouble knowing what the molar mass is can you remind me?

Answer 12

Add up the individual molar masses in the right proportions

molar mass of iron = 55.8 g/mol

molar mas of oxygen = 16 g/mol

there are 2 Fe, 3 O,

molar mass of \(Fe_2O_3\)= 2(55.8 g/mol)+3(16 g/mol)= 159.6 g/mol

Answer 13

And just the mass is the one in the given right?

Answer 14

yes 7.65 grams Fe2O3

you need to find the moles of CO as well

Answer 15

CO is 58.9. and its that I am getting a big decimal number and I feel that it isn't right when I divide the mass by the molar mass

Answer 16

I am getting 0.02976

Answer 17

@aaronq

Answer 18

CO is 12+16=28 g/mol

Answer 19

so what are the moles of each?

Answer 20

So why do I need to find the CO?

Answer 21

Because you need to determine which of your starting products will run out first.

Answer 22

So I am assuming I need to do the same to CO that I did to Fe2O3 so I got 0.27321

Answer 23

Now to find the limiting reactant, we divide the moles of each reactant by their corresponding stoichiometric coefficient. And compare, the one with less moles is the limiting reactant.

Answer 24

I think that the limiting reactant is either CO or Fe

Answer 25

we need to know which one though.

Answer 26

I think the limiting reactant is CO

Answer 27

Am I right?

Answer 28

sorry dude, the site is freezing on me

Answer 29

what were your values?

Answer 30

What do you mean "values"?

Answer 31

".. divide the moles of each reactant by their corresponding stoichiometric coefficient."

Answer 32

So does that mean I divide CO by the coefficient which is 3? and do the same for Fe?

Answer 33

yes

Answer 34

So I have 0.27321/3 which gives me 0.09107

Answer 35

so whats the limiting reactant?

Answer 36

and for Fe I think it is 0.04793/2 which gives me 0.023966. The reason I am typing everything I am doing is because I want to make sure I am doing it right.

Answer 37

And the limiting reactant is the one with the lowest right?

Answer 38

Okay you went ahead and finished the problem.

Now find the mass of Fe with those moles

Answer 39

sorry i thought we were a step behind

Answer 40

Okay Now I am confused at which step we are at lol.

Answer 41

You said "and for Fe.. "

Did you find the moles of Fe? are you at that step?

I thought we were still working with \(Fe_2O_3\)

Answer 42

Oh yeah I did jumped ahead I am sorry I forgot we were doing reactants

Answer 43

SO you have the moles of Fe? what is it?

Answer 44

So Like you said Fe2O3 has 1 coefficient

Answer 45

No I don't It's that I got confused in the middle of it sorry I am still at Fe2O3

Answer 46

Okay, so we were finding the limiting reactant. Did you determine which had less moles after dividing by their coefficients?

Answer 47

I think I did and I think the one with the less moles was Fe2O3 has less moles

Answer 48

okay, good stuff. Now we make a ratio of the moles of the limiting reactant (\(Fe_2O_3\) in this case) and the moles of what we want to find (Fe in this case) with their coefficients (like in the previous problem).

Answer 49

\(\sf \dfrac{moles~of ~Fe_2O_3}{Fe_2O_3's~ coefficient}=\dfrac{moles~of~Fe}{Fe's ~coefficient}\)

Answer 50

Can you help me find the moles of Fe please?

Answer 51

wait hold on

Answer 52

i gotta get going. After you find the moles of Fe, you need to convert them to mass. (This is the theoretical yield.)

Use: \(n=\dfrac{m}{M}\)


After, to find the percent yield, use:

\(\sf\% ~yield=\dfrac{actual~yield}{theoretical~yield}*100\%\)