Understanding the Limiting Reactant
In chemical reactions, the limiting reactant is the substance that is completely consumed first, thereby determining the maximum amount of product that can be formed. To find it, we must compare the mole ratios provided by the balanced equation to the actual moles available.
The Balanced Equation
$$Fe_{2}O_{3} + 3CO \rightarrow 2Fe + 3CO_{2}$$
Step-by-Step Solution
1. Calculate Molar Masses:
- $Fe_{2}O_{3}: (2 \times 56) + (3 \times 16) = 112 + 48 = 160 \text{ g/mol}$
- $CO: 12 + 16 = 28 \text{ g/mol}$
2. Calculate Initial Moles:
- $Moles_{Fe_{2}O_{3}} = \frac{130 \text{ g}}{160 \text{ g/mol}} = 0.8125 \text{ mol}$
- $Moles_{CO} = \frac{50 \text{ g}}{28 \text{ g/mol}} \approx 1.786 \text{ mol}$
i. Finding the Limiting Reactant: From the equation, $1 \text{ mol of } Fe_{2}O_{3}$ requires $3 \text{ mol of } CO$. We need: $0.8125 \times 3 = 2.4375 \text{ mol of } CO$. Since we only have $1.786 \text{ mol of } CO$, CO is the limiting reactant.
ii. Mass of Iron Produced: Based on $CO$ (the limiting reagent), the ratio of $CO:Fe$ is $3:2$. $Moles_{Fe} = \frac{2}{3} \times 1.786 = 1.191 \text{ mol}$. $Mass_{Fe} = 1.191 \text{ mol} \times 56 \text{ g/mol} = 66.696 \text{ g}$.
iii. Unreacted Reactant Left: $Moles_{Fe_{2}O_{3}} \text{ used} = \frac{1}{3} \times 1.786 = 0.595 \text{ mol}$. $Moles_{Fe_{2}O_{3}} \text{ remaining} = 0.8125 - 0.595 = 0.2175 \text{ mol}$.
iv. Volume of $CO_{2}$ Produced: Ratio of $CO:CO_{2}$ is $3:3$, so $1.786 \text{ mol of } CO_{2}$ is produced. Using the ideal gas law: $PV = nRT$ $V = \frac{nRT}{P} = \frac{1.786 \times 0.0821 \times 298}{2} = 21.84 \text{ Liters}$.