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Certainly! Here's a stoichiometry question for you:

**Question:**
When solid iron (III) oxide (Fe2O3) is reduced with carbon monoxide gas (CO) in the presence of excess carbon (C) to produce iron (Fe), carbon dioxide gas (CO2), and carbon monoxide gas (CO) as byproducts according to the following balanced equation:

\[ Fe2O3(s) + 3CO(g) + 3C(s) \rightarrow 2Fe(s) + 3CO2(g) + 3CO(g) \]

1. If 200 grams of iron (III) oxide (Fe2O3) is mixed with 130 grams of carbon monoxide (CO) and excess carbon (C), calculate:
a) The limiting reactant

b) The theoretical yield of iron (Fe) in grams
c) The percentage yield of iron if the actual yield obtained is 145 grams

Feel free to work through this stoichiometry question and let me know if you need any assistance with the calculations or concepts!
Well 💩 my brain has decided to go on vacation

Give me the answers and working out
 
Well 💩 my brain has decided to go on vacation

Give me the answers and working out
No worries! Let's work through the stoichiometry question together:

**Given:**
- Iron (III) oxide (Fe2O3) = 200 grams
- Carbon Monoxide (CO) = 130 grams

**Molar Mass:**
- Fe2O3 = 2*(55.85 g/mol) + 3*(16.00 g/mol) = 159.7 g/mol
- CO = 12.01 g/mol + 15.999 g/mol = 28.01 g/mol

**Calculations:**
**a) The limiting reactant:**

1. Calculate the moles of Fe2O3:
\[ \text{Moles of Fe2O3} = \frac{200 \text{ g}}{159.7 \text{ g/mol}} \approx 1.25 \text{ mol} \]

2. Calculate the moles of CO:
\[ \text{Moles of CO} = \frac{130 \text{ g}}{28.01 \text{ g/mol}} \approx 4.64 \text{ mol} \]

3. Determine the limiting reactant by comparing the mole ratios from the balanced equation:
\[ \frac{\text{Moles of Fe2O3}}{1} = \frac{\text{Moles of CO}}{3} \]
Since Fe2O3:CO ratio is 1:3, CO is the limiting reactant in this case.

**b) The theoretical yield of iron (Fe) in grams:**

1. Calculate the theoretical yield of Fe (using the limiting reactant - CO):
From the balanced equation, 3 moles of CO produce 2 moles of Fe.
\[ \text{Moles of Fe} = \frac{4.64 \text{ mol CO} \times 2 \text{ mol Fe}}{3 \text{ mol CO}} \approx 3.09 \text{ mol Fe} \]

2. Calculate the theoretical yield in grams:
\[ \text{Theoretical yield of Fe} = 3.09 \text{ mol Fe} \times 55.85 \text{ g/mol} \approx 172.8 \text{ g Fe} \]

**c) The percentage yield of iron if the actual yield obtained is 145 grams:**

1. Calculate the percentage yield:
\[ \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \]
\[ \text{Percentage Yield} = \frac{145 \text{ g}}{172.8 \text{ g}} \times 100\% \approx 83.9\% \]

Therefore:
a) The limiting reactant is CO
b) The theoretical yield of iron (Fe) is approximately 172.8 grams
c) The percentage yield of iron is approximately 83.9%

I hope this helps! If you have any more questions or need further clarification, feel free to ask.
 
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No worries! Let's work through the stoichiometry question together:

**Given:**
- Iron (III) oxide (Fe2O3) = 200 grams
- Carbon Monoxide (CO) = 130 grams

**Molar Mass:**
- Fe2O3 = 2*(55.85 g/mol) + 3*(16.00 g/mol) = 159.7 g/mol
- CO = 12.01 g/mol + 15.999 g/mol = 28.01 g/mol

**Calculations:**
**a) The limiting reactant:**

1. Calculate the moles of Fe2O3:
\[ \text{Moles of Fe2O3} = \frac{200 \text{ g}}{159.7 \text{ g/mol}} \approx 1.25 \text{ mol} \]

2. Calculate the moles of CO:
\[ \text{Moles of CO} = \frac{130 \text{ g}}{28.01 \text{ g/mol}} \approx 4.64 \text{ mol} \]

3. Determine the limiting reactant by comparing the mole ratios from the balanced equation:
\[ \frac{\text{Moles of Fe2O3}}{1} = \frac{\text{Moles of CO}}{3} \]
Since Fe2O3:CO ratio is 1:3, CO is the limiting reactant in this case.

**b) The theoretical yield of iron (Fe) in grams:**

1. Calculate the theoretical yield of Fe (using the limiting reactant - CO):
From the balanced equation, 3 moles of CO produce 2 moles of Fe.
\[ \text{Moles of Fe} = \frac{4.64 \text{ mol CO} \times 2 \text{ mol Fe}}{3 \text{ mol CO}} \approx 3.09 \text{ mol Fe} \]

2. Calculate the theoretical yield in grams:
\[ \text{Theoretical yield of Fe} = 3.09 \text{ mol Fe} \times 55.85 \text{ g/mol} \approx 172.8 \text{ g Fe} \]

**c) The percentage yield of iron if the actual yield obtained is 145 grams:**

1. Calculate the percentage yield:
\[ \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \]
\[ \text{Percentage Yield} = \frac{145 \text{ g}}{172.8 \text{ g}} \times 100\% \approx 83.9\% \]

Therefore:
a) The limiting reactant is CO
b) The theoretical yield of iron (Fe) is approximately 172.8 grams
c) The percentage yield of iron is approximately 83.9%

I hope this helps! If you have any more questions or need further clarification, feel free to ask.
Give me a simpler question, more similar to the ones u gave me earlier in our conversation a few days ago
 

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