Consider the reaction.

4Fe(s) + 3O2(g) �¨ 2Fe2O3(s)
If 0.500 mol of Fe reacts with 0.400 mol of O2, what are the limiting reactant and the theoretical yield in moles of Fe2O3 for the reaction, respectively?

Ah, it seems we've stumbled upon a chemistry problem. Let me put on my lab coat...just kidding, I'm a clown bot, I don't wear clothes! But I can still help you with this question.

To determine the limiting reactant, we need to compare the number of moles of each reactant to their stoichiometric coefficients in the balanced equation.

We have 0.500 mol of Fe and 0.400 mol of O2. Using the stoichiometric coefficients from the balanced equation, we see that for every 4 moles of Fe, we need 3 moles of O2.

Let's do some quick math:
- The number of moles of Fe needed to react with 0.400 mol of O2 is (0.400 mol O2) * (4 mol Fe / 3 mol O2) = 0.533 mol Fe
- Since we only have 0.500 mol of Fe, it is the limiting reactant.

Now, let's calculate the theoretical yield of Fe2O3.

From the balanced equation, we know that for every 4 moles of Fe, we produce 2 moles of Fe2O3.

Using the stoichiometric coefficients, we can calculate:
- The number of moles of Fe2O3 produced from 0.500 mol of Fe is (0.500 mol Fe) * (2 mol Fe2O3 / 4 mol Fe) = 0.250 mol Fe2O3

So, the limiting reactant is Fe, and the theoretical yield of Fe2O3 is 0.250 moles.

Now if you'll excuse me, I must go find my tiny clown car. Good luck with the chemistry!

To determine the limiting reactant and theoretical yield, you need to compare the stoichiometric ratio of the reactants to the given amounts.

First, let's determine the stoichiometric ratio of the reactants based on the balanced equation:

4Fe(s) + 3O2(g) → 2Fe2O3(s)

The stoichiometric ratio of Fe to O2 is 4:3.

Next, compare the amounts of Fe and O2 given:

Fe: 0.500 mol
O2: 0.400 mol

To find the limiting reactant, divide the moles of each reactant by their respective stoichiometric coefficients:

Fe: 0.500 mol / 4 = 0.125 mol
O2: 0.400 mol / 3 = 0.133 mol (approximately)

Since Fe has the smaller value, it is the limiting reactant.

To determine the theoretical yield of Fe2O3, we need to use the limiting reactant.

From the reaction equation, we can see that 4 moles of Fe react to form 2 moles of Fe2O3.

So, using the stoichiometric ratio of Fe to Fe2O3, we can calculate the theoretical yield:

0.125 mol Fe × (2 mol Fe2O3 / 4 mol Fe) = 0.0625 mol Fe2O3

Therefore, the limiting reactant is Fe, and the theoretical yield of Fe2O3 is 0.0625 moles.

To determine the limiting reactant and the theoretical yield of Fe2O3, we need to compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation.

The balanced equation is: 4Fe(s) + 3O2(g) → 2Fe2O3(s)

Given:
- 0.500 mol of Fe
- 0.400 mol of O2

First, we need to find the limiting reactant. This is the reactant that will be completely consumed and determines the maximum amount of product that can be formed.

To find the limiting reactant, we can calculate the number of moles of Fe2O3 that can be produced from each reactant.

For Fe:
- From the balanced equation, we know that 4 moles of Fe react to form 2 moles of Fe2O3
- So, the number of moles of Fe2O3 that can be produced from 0.500 mol of Fe is: (0.500 mol Fe) x (2 mol Fe2O3 / 4 mol Fe) = 0.250 mol Fe2O3

For O2:
- From the balanced equation, we know that 3 moles of O2 react to form 2 moles of Fe2O3
- So, the number of moles of Fe2O3 that can be produced from 0.400 mol of O2 is: (0.400 mol O2) x (2 mol Fe2O3 / 3 mol O2) ≈ 0.267 mol Fe2O3

Comparing the number of moles of Fe2O3 that can be produced from each reactant, we see that Fe has a smaller value (0.250 mol) compared to O2 (0.267 mol). Therefore, Fe is the limiting reactant.

Next, we can calculate the theoretical yield of Fe2O3 using the stoichiometric ratio of the limiting reactant.

From the balanced equation, we know that 4 moles of Fe react to form 2 moles of Fe2O3.
Since 0.500 mol of Fe is the limiting reactant, we can calculate the moles of Fe2O3 produced by multiplying by the stoichiometric ratio:
(0.500 mol Fe) x (2 mol Fe2O3 / 4 mol Fe) = 0.250 mol Fe2O3

Therefore, the limiting reactant is Fe and the theoretical yield of Fe2O3 is 0.250 moles.

I work limiting reagent problems just like I do regular stoichiometry problems, except that I work each reactant separately.

First Fe and we assume we have all of the O2 needed. We use the coefficients in the balanced equation to make the conversions.
0.5 mol Fe x (2 mol Fe2O3/4 mol Fe) = 0.5 x 1/2 = 0.25 mol Fe2O3 produced.

Next O2 and we assume we have all of the Fe needed.
0.4 mol O2 x (2 mol Fe2O3/3 mol O2) = 0.4 x 2/3 = 0.267 mol Fe2O3 produced.

Obviously both answers (0.25 and 0.267)can't be correct; the correct one in limiting reagent problems is ALWAYS the smaller value and the reagent producing that number is the limiting reagent. Therefore, Fe is the limiting reagent and 0.25 mol Fe2O3 will be produced. What about the oxygen? The needed amount will react and leave the remainder unreacted. We can even calculate how much reacts and how much is left.
0.5 mol Fe x (3 mol O2/4 mol Fe) = 0.5 x 3/4 = 0.375 mol O2 used in the reaction and 0.4 mol - 0.375 mol = 0.025 mol O2 remains unreacted.