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?
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
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.
I want to Convert/ Calculate the Fe % to Fe2O3 %
To determine the limiting reactant, you need to compare the number of moles of each reactant used in the reaction. The reactant that produces the smallest amount of product is the limiting reactant. Here's how you can find it:
Step 1: Find the molar masses of the reactants.
The molar mass of Fe2O3 (ferric oxide) = 2 * (55.845 g/mol) + 3 * (16.00 g/mol) = 159.69 g/mol
The molar mass of CO (carbon monoxide) = 12.01 g/mol + 16.00 g/mol = 28.01 g/mol
Step 2: Convert the masses of each reactant to moles.
Number of moles of Fe2O3 = mass / molar mass = 0.5300 g / 159.69 g/mol = 0.0033188 mol
Number of moles of CO = mass / molar mass = 0.3701 g / 28.01 g/mol = 0.013209 mol
Step 3: Use the stoichiometry of the balanced equation to determine the theoretical amount of product formed by each reactant.
According to the balanced equation, the stoichiometric ratio between Fe2O3 and Fe is 1:2. This means that for every 1 mole of Fe2O3, 2 moles of Fe are produced.
The stoichiometric ratio between CO and Fe is 1:1. This means that for every 1 mole of CO, 1 mole of Fe is produced.
Theoretical moles of Fe produced from Fe2O3 = 0.0033188 mol * 2 = 0.0066376 mol
Theoretical moles of Fe produced from CO = 0.013209 mol
Since we have excess CO, the limiting reactant is Fe2O3 because it produces a smaller amount of Fe.
A. The limiting reactant is Fe2O3.
B. To determine the mass of the other reactant in excess, you need to subtract the amount of the limiting reactant used from the total amount of the other reactant.
Mass of CO in excess = total mass of CO - mass of CO used
= 0.3701 g - (0.0033188 mol * 28.01 g/mol)
= 0.2781 g
C. The mass of iron produced can be calculated from the moles of Fe produced by the limiting reactant. Since the stoichiometric ratio is 1:1 between Fe2O3 and Fe, the moles of Fe produced is equal to the moles of Fe2O3 used.
Mass of iron produced = moles of Fe2O3 used * molar mass of Fe
= 0.0033188 mol * 55.845 g/mol
= 0.1853 g
Therefore, the mass of iron produced is 0.1853 g.