Questions: Part C: Given the reaction quantities in the image above, how many ammo molecules can be created? choose your answer... ✓ molecule(s) Part D: How many molecules of excess reactant remain AFTER the reaction to completion? □ 2 molecule(s) 1 point Consider the following balanced chemical reaction for Questions 7-9, Fe2O3(s) + 3 CO(g) -> 2 Fe(s) + 3 CO2(g) If 0.22 mol of Fe2O3 is reacted with 5.5 mol of CO , what is the limiting reactant? Fe2O3 CO Fe CO2

Solution Steps

The balanced chemical equation given is: \[ \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s}) + 3 \mathrm{CO}(\mathrm{g}) \rightarrow 2 \mathrm{Fe}(\mathrm{s}) + 3 \mathrm{CO}_{2}(\mathrm{g}) \]

We are given:

• 0.22 mol of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\)
• 5.5 mol of \(\mathrm{CO}\)

To find the limiting reactant, we need to compare the mole ratio of the reactants with the coefficients in the balanced equation.

From the balanced equation:

• 1 mol of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) reacts with 3 mol of \(\mathrm{CO}\)

For 0.22 mol of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\): \[ 0.22 \, \text{mol} \, \mathrm{Fe}_{2} \mathrm{O}_{3} \times \frac{3 \, \text{mol} \, \mathrm{CO}}{1 \, \text{mol} \, \mathrm{Fe}_{2} \mathrm{O}_{3}} = 0.66 \, \text{mol} \, \mathrm{CO} \]

We have 5.5 mol of \(\mathrm{CO}\) available, which is more than the 0.66 mol required to react with 0.22 mol of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\).

Since 0.22 mol of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) requires only 0.66 mol of \(\mathrm{CO}\) and we have 5.5 mol of \(\mathrm{CO}\), \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is the limiting reactant.

Final Answer