The reduction of iron (III) oxide creates the following reaction: Fe 2 O 3 (s) + 3 H 2 (g) → 2 Fe (s) + 3 H 2 O (g) If the above reaction only went to 75% completion, how many moles of Fe 2 O 3 were require to produce 0.850 moles of Fe?

In this practice question it says, the reduction of iron 3 oxide creates the following reaction. We have 1 mole of iron 3 solid reacting with 3 moles of hydrogen gas to produce 2 moles of iron solid plus 3 moles of water vapor. Here it says if the above reaction only went to 75% completion, how many moles of iron 3 oxide were required to produce 0.8 50 moles of product. Now if they're telling us how much product was made, produced, isolated, the only way we would know that is if we did the experiment in real life. And so this would have to represent my actual yield. Because remember, your actual yield is what you make in real life. Your theoretical yield is what you calculate down on paper. Now here, they're giving us a percentage, so this has to be my percent yield. So then it becomes percent yield equals actual yield over theoretical yield times 100. We plug in our numbers, 75% here. Our actual yield is 0.850 moles of iron. Remember that the units for actual yield and theoretical yield must match, so my theoretical yield would be x moles of iron. So this becomes at this point a way of isolating moles of iron. Alright. So now we're going to multiply both sides here by x, so this becomes 75x equals 0.850 times 100. Divide both sides here by 75, and that'll isolate our x variable which we just said represents our moles of iron. So x equals 1.133 moles of iron. Now this is not the answer we're looking for. We're not looking for moles of iron. We're looking for moles of iron 3 oxide. But remember, iron and iron 3 oxide are connected to each other through the chemical equation, and we can go from moles of 1 to moles of the other by doing a mole to mole comparison. So here we're gonna put moles of iron on the bottom and here we're gonna put moles of iron 3 oxide on the top. Remember, to go do a mole to mole comparison, we have to look at the coefficients in the balanced equation, and it is a 1 to 2 relationship. At the end, we see that we have 0.5665 moles of iron 3 oxide, And realize this represents the moles of the reactant that were required to produce 0.850 moles of our product.

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1. Matter and Measurements4h 31m
2. Atoms and the Periodic Table5h 23m
3. Ionic Compounds2h 18m
4. Molecular Compounds2h 18m
5. Classification & Balancing of Chemical Reactions3h 17m
6. Chemical Reactions & Quantities2h 27m
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6. Chemical Reactions & Quantities

Percent Yield

GOB Chemistry

6. Chemical Reactions & Quantities

Percent Yield

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Multiple Choice

The reduction of iron (III) oxide creates the following reaction:
Fe₂O₃ (s) + 3 H₂ (g) → 2 Fe (s) + 3 H₂O (g)
If the above reaction only went to 75% completion, how many moles of Fe₂O₃ were require to produce 0.850 moles of Fe?

0.963 mol

2.27 mol

0.425 mol

0.567 mol

1.70 mol

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Verified step by step guidance

Understand the stoichiometry of the reaction: Fe2O3 reacts with H2 to produce Fe and H2O. The balanced equation is Fe2O3 + 3 H2 → 2 Fe + 3 H2O.

Identify the mole ratio between Fe2O3 and Fe from the balanced equation. For every 1 mole of Fe2O3, 2 moles of Fe are produced.

Since the reaction goes to 75% completion, calculate the effective moles of Fe produced. Multiply the given moles of Fe (0.850 moles) by 100/75 to find the moles of Fe that would be produced if the reaction went to full completion.

Use the mole ratio to find the moles of Fe2O3 required for the full completion moles of Fe. Divide the moles of Fe (from step 3) by 2 (from the stoichiometry) to find the moles of Fe2O3 needed.

Consider the options provided and identify the correct number of moles of Fe2O3 that matches your calculation from step 4.