In an experiment 0.5300 g of ferric oxide is reacted with 0.3701 g of carbon monoxide gas according to the equation:
Fe2O3 + CO ==> Fe + CO2.
A. What is the limiting reactant?
B. What mass of the other reactant is in exess?
C. What mass of iron is produced?
6 answers
This is a long problem. Tell us what you don't understand about it. I don't mind helping your through it but I don't want to waste my time going over something you already know how to do.
i have no idea how to do it
at all
ive been absent for 3 weeks
at all
ive been absent for 3 weeks
Do you know how to do stoichiometry?
kind of yeah
I will save some time by estimating molar masses and using a minimum of numbers. You need to go back through it and redo them exactly. Check all my arithmetic procedures, too.
Step 1. Write and balance the equation.
Fe2O3 + 3CO ==> 2Fe + 3CO2
Step 2. Convert what you have into mols. In limiting reagent problems you always have at least TWO substances given to you. They are CO and Fe2O3 in this problem. Remember mols = grams/molar mass.
2a. Convert CO to mols.
0.3701/28 = 0.013 mols CO
2b. Convert Fe2O3 to mols.
0.5300/160 = 0.0033
Step 3. Using the coefficients in the balanced equation, convert BOTH 2a mols and 2b mols to mols Fe (you could chose ANY product but since part c asks for Fe we may as well save a little time.).
3a. First, convert mols CO to mols Fe.
mols Fe = mols CO x (2 mols Fe/3 mols CO) = 0.013 x (2/3) = 0.0087 mols Fe.
3b. Now convert Fe2O3 to mols Fe.
mols Fe = mols Fe2O3 x (2 mols Fe/1 mol Fe2O3) = 0.033 x (2/1) = 0.0o66
3c. Obviously, both answers CAN'T be right. The correct answer is ALWAYS the smaller of the two, in this case 0.0066
3d. What produced the 0.0066? That is the 0.5300 g Fe2O3; therefore, THAT is the limiting reagent (reactant) (answer to part a) and CO is the OTHER reagent (reactant).
Step 4. Now that we know how many mols Fe are produced, we convert that to grams (answer to part c) by
grams = mols x atomic mass = 0.0066 x 55.8 = 0.37 g Fe.
If you will get these four steps down, they will work almost ANY stoichiometry problem OR limiting reagent problem.
I won't go through the part b but here is how you do it.
It now becomes a separate problem. Here is the way I would state it. How many grams of CO will be consumed (and how much will remain unreacted) if we react 0.5300 g Fe2O3 with 0.3701 g CO.
You already have step 1, the equation, and step 2, mols Fe2O3. Just go through step 3 (you can skip over the 3b,3c and 3d) and convert mols Fe2O3 to mols CO, then convert mols CO to grams (as in step 4), then subtract this number from the 0.3701 you started with and you will have the amount of CO remaining unreacted.
Post your work if you get stuck but be sure and tell me exactly what you don't understand about the next step where you are stuck. I hope this helps. By the way, there are easier ways to do limiting regent problems but I think this is the easiest way to explain it.
Step 1. Write and balance the equation.
Fe2O3 + 3CO ==> 2Fe + 3CO2
Step 2. Convert what you have into mols. In limiting reagent problems you always have at least TWO substances given to you. They are CO and Fe2O3 in this problem. Remember mols = grams/molar mass.
2a. Convert CO to mols.
0.3701/28 = 0.013 mols CO
2b. Convert Fe2O3 to mols.
0.5300/160 = 0.0033
Step 3. Using the coefficients in the balanced equation, convert BOTH 2a mols and 2b mols to mols Fe (you could chose ANY product but since part c asks for Fe we may as well save a little time.).
3a. First, convert mols CO to mols Fe.
mols Fe = mols CO x (2 mols Fe/3 mols CO) = 0.013 x (2/3) = 0.0087 mols Fe.
3b. Now convert Fe2O3 to mols Fe.
mols Fe = mols Fe2O3 x (2 mols Fe/1 mol Fe2O3) = 0.033 x (2/1) = 0.0o66
3c. Obviously, both answers CAN'T be right. The correct answer is ALWAYS the smaller of the two, in this case 0.0066
3d. What produced the 0.0066? That is the 0.5300 g Fe2O3; therefore, THAT is the limiting reagent (reactant) (answer to part a) and CO is the OTHER reagent (reactant).
Step 4. Now that we know how many mols Fe are produced, we convert that to grams (answer to part c) by
grams = mols x atomic mass = 0.0066 x 55.8 = 0.37 g Fe.
If you will get these four steps down, they will work almost ANY stoichiometry problem OR limiting reagent problem.
I won't go through the part b but here is how you do it.
It now becomes a separate problem. Here is the way I would state it. How many grams of CO will be consumed (and how much will remain unreacted) if we react 0.5300 g Fe2O3 with 0.3701 g CO.
You already have step 1, the equation, and step 2, mols Fe2O3. Just go through step 3 (you can skip over the 3b,3c and 3d) and convert mols Fe2O3 to mols CO, then convert mols CO to grams (as in step 4), then subtract this number from the 0.3701 you started with and you will have the amount of CO remaining unreacted.
Post your work if you get stuck but be sure and tell me exactly what you don't understand about the next step where you are stuck. I hope this helps. By the way, there are easier ways to do limiting regent problems but I think this is the easiest way to explain it.
I want to Convert/ Calculate the Fe % to Fe2O3 %
24 answers