Calculate the mass of iron produced from 20 g of 50% pure Fe2O3 and 9 g of CO, given the reaction: Fe2O3 + 3CO → 2Fe + 3CO2.

Step 1: Calculate the actual mass and moles of reactants.
The balanced chemical equation is:

$$\text{Fe}_2\text{O}_3 + 3\text{CO} \longrightarrow 2\text{Fe} + 3\text{CO}_2$$

Given: 20 g of 50% pure $\text{Fe}_2\text{O}_3$ and 9 g of $\text{CO}$.

First, calculate the actual mass of pure $\text{Fe}_2\text{O}_3$:

$$\text{Mass of pure Fe}_2\text{O}_3 = 20 \text{ g} \times 0.50 = 10 \text{ g}$$

Next, calculate the molar masses of the reactants:

$$\text{Molar mass of Fe}_2\text{O}_3 = (2 \times 55.845) + (3 \times 15.999) = 159.687 \text{ g/mol}$$ $$\text{Molar mass of CO} = 12.011 + 15.999 = 28.010 \text{ g/mol}$$

Now, calculate the moles of each reactant:

$$\text{Moles of Fe}_2\text{O}_3 = \frac{10 \text{ g}}{159.687 \text{ g/mol}} \approx 0.06262 \text{ mol}$$ $$\text{Moles of CO} = \frac{9 \text{ g}}{28.010 \text{ g/mol}} \approx 0.32131 \text{ mol}$$

Step 2: Determine the limiting reactant.
From the balanced equation, 1 mole of $\text{Fe}_2\text{O}_3$ reacts with 3 moles of $\text{CO}$.

To react completely with $0.06262 \text{ mol}$ of $\text{Fe}_2\text{O}_3$, the moles of $\text{CO}$ required would be:

$$\text{Moles of CO required} = 0.06262 \text{ mol Fe}_2\text{O}_3 \times \frac{3 \text{ mol CO}}{1 \text{ mol Fe}_2\text{O}_3} = 0.18786 \text{ mol CO}$$

Since we have $0.32131 \text{ mol}$ of $\text{CO}$ (which is more than $0.18786 \text{ mol}$ required), $\text{Fe}_2\text{O}_3$ is the limiting reactant.

a) i) The limiting reactant is: $\boxed{\text{Fe}_2\text{O}_3}$

Step 3: Calculate the moles of unreacted reactant left over.
The excess reactant is $\text{CO}$.
Moles of $\text{CO}$ consumed in the reaction:

$$\text{Moles of CO consumed} = 0.18786 \text{ mol}$$

Moles of $\text{CO}$ left over:

$$\text{Moles of CO left over} = \text{Initial moles of CO} - \text{Moles of CO consumed}$$ $$\text{Moles of CO left over} = 0.32131 \text{ mol} - 0.18786 \text{ mol} = 0.13345 \text{ mol}$$

a) ii) Moles of unreacted reactant left over: $\boxed{0.133 \text{ mol CO}}$

Step 4: Calculate the number of moles of CO consumed in the reaction.
This was calculated in Step 3.

a) iii) Number of moles of CO consumed: $\boxed{0.188 \text{ mol CO}}$

Step 5: Calculate the mass of NaOH required to absorb all $\text{CO}_2$ produced.
First, calculate the moles of $\text{CO}_2$ produced. Since $\text{Fe}_2\text{O}_3$ is the limiting reactant:

$$\text{Moles of CO}_2 \text{ produced} = 0.06262 \text{ mol Fe}_2\text{O}_3 \times \frac{3 \text{ mol CO}_2}{1 \text{ mol Fe}_2\text{O}_3} = 0.18786 \text{ mol CO}_2$$

The reaction between $\text{CO}_2$ and $\text{NaOH}$ is:

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