Let's start by analyzing the balanced chemical reaction and then proceed with the stoichiometric calculations.

The balanced equation for the reaction between magnesium (Mg) and iron(III) chloride (FeCl₃) is:

$3\text{Mg(s)} + 2\text{FeCl}_3\text{(aq)} \rightarrow 3\text{MgCl}_2\text{(aq)} + 2\text{Fe(s)}$

Step 1: Calculate the moles of magnesium (Mg).

We know that the mass of magnesium is given as 3.75 g. Using the molar mass of magnesium ($M_{Mg} = 24.305 \, \text{g/mol}$), we can find the moles of magnesium.

$n_{\text{Mg}} = \frac{3.75 \, \text{g}}{24.305 \, \text{g/mol}} = 0.1542 \, \text{mol}$

Step 2: Calculate the moles of iron(III) chloride (FeCl₃).

The concentration of the FeCl₃ solution is given as 1.23 M, and the volume is 100.0 mL or 0.100 L. The moles of FeCl₃ can be calculated using the formula:

$n_{\text{FeCl}_3} = M \times V = 1.23 \, \text{mol/L} \times 0.100 \, \text{L} = 0.123 \, \text{mol}$

Step 3: Determine the limiting reactant.

The balanced equation shows that 3 moles of Mg react with 2 moles of FeCl₃. We now compare the molar ratio of the reactants to determine the limiting reactant.

$\text{Mole ratio of Mg to FeCl}_3 = \frac{n_{\text{Mg}}}{n_{\text{FeCl}_3}} = \frac{0.1542 \, \text{mol}}{0.123 \, \text{mol}} = 1.25$

The stoichiometric ratio from the balanced equation is:

$\frac{3 \, \text{mol} \, \text{Mg}}{2 \, \text{mol} \, \text{FeCl}_3} = 1.5$

Since the actual mole ratio (1.25) is less than the stoichiometric ratio (1.5), this means that FeCl₃ is in excess and Mg is the limiting reactant.

Step 4: Calculate the moles of iron (Fe) produced.

From the balanced equation, 3 moles of Mg produce 2 moles of Fe. Therefore, the moles of Fe produced are:

$n_{\text{Fe}} = \frac{2}{3} \times n_{\text{Mg}} = \frac{2}{3} \times 0.1542 \, \text{mol} = 0.1028 \, \text{mol}$

Step 5: Calculate the mass of iron (Fe) produced.

The molar mass of iron is $M_{\text{Fe}} = 55.845 \, \text{g/mol}$. Using this, the mass of iron produced is:

$m_{\text{Fe}} = n_{\text{Fe}} \times M_{\text{Fe}} = 0.1028 \, \text{mol} \times 55.845 \, \text{g/mol} = 5.74 \, \text{g}$

Final Answer:

The mass of iron prepared by the reaction is 5.74 g.

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Do you want more details or have any questions about the process? Here are 5 additional related questions you might want to explore:

1. What is the concept of limiting reactants in stoichiometry?
2. How is molarity used in calculating the amount of a substance in a solution?
3. Can you explain why magnesium is the limiting reactant in this case?
4. How would the result change if the amount of magnesium were doubled?
5. How does the conservation of mass apply in this reaction?

Tip: Always balance chemical equations first before performing stoichiometric calculations to ensure accuracy!