Question

How many grams of iron(III) sulfide form when \(62.0 \mathrm{~mL}\) of \(0.135 \mathrm{M}\) iron(III) chloride reacts with \(45.0 \mathrm{~mL}\) of \(0.285 \mathrm{M}\) calcium sulfide?

Short answer

1.272 grams of Fe\textsubscript{2}S\textsubscript{3} are formed.

Step-by-step solution

Write the balanced chemical equation

The reaction between iron(III) chloride and calcium sulfide forms iron(III) sulfide and calcium chloride. The balanced chemical equation is: \[2 \text{FeCl}_3 (aq) + 3 \text{CaS} (aq) \rightarrow \text{Fe}_2\text{S}_3 (s) + 3 \text{CaCl}_2 (aq)\]

Calculate moles of FeCl\textsubscript{3}

Use the molarity and volume of FeCl\textsubscript{3} to find the moles. \[ \text{Moles of FeCl}_3 = 0.135 \text{ M} \times 62.0 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.00837 \text{ mol} \]

Calculate moles of CaS

Use the molarity and volume of CaS to find the moles. \[ \text{Moles of CaS} = 0.285 \text{ M} \times 45.0 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.012825 \text{ mol} \]

Determine the limiting reactant

Compare the mole ratio with the balanced equation. \[2 \text{ moles of FeCl}_3 \text{ reacts with } 3 \text{ moles of CaS} \] Therefore, \[ \frac{0.00837 \text{ mol FeCl}_3}{2} = 0.004185 \text{ mol} \] and \[ \frac{0.012825 \text{ mol CaS}}{3} = 0.004275 \text{ mol} \] Since 0.00837 mol FeCl\textsubscript{3} is less than 0.012825 mol CaS (adjusted by the mole ratio), FeCl\textsubscript{3} is the limiting reactant.

Calculate moles of Fe\textsubscript{2}S\textsubscript{3} formed

Use the moles of the limiting reactant to determine the moles of Fe\textsubscript{2}S\textsubscript{3} produced. The balanced equation shows 2 mol FeCl\textsubscript{3} produces 1 mol Fe\textsubscript{2}S\textsubscript{3}, so: \[ \text{Moles of Fe}_2\text{S}_3 = \frac{0.00837 \text{ mol FeCl}_3}{2} = 0.004185 \text{ mol} \]

Calculate grams of Fe\textsubscript{2}S\textsubscript{3}

Finally, use the molar mass of Fe\textsubscript{2}S\textsubscript{3} (207.87 g/mol for Fe and 32.06 g/mol for S). The molar mass of Fe\textsubscript{2}S\textsubscript{3} is \[2 \times 55.85 + 3 \times 32.06 = 207.87 + 96.18 = 303.91 \text{ g/mol} \] Therefore, \[ \text{Mass of Fe}_2\text{S}_3 = 0.004185 \text{ mol} \times 303.91 \text{ g/mol} = 1.272 \text{ g} \]

Key concepts

balanced chemical equation

A balanced chemical equation represents the reactants and products in a chemical reaction with their respective quantities. This is crucial because it ensures the law of conservation of mass is followed. In our problem, we have iron(III) chloride reacting with calcium sulfide to produce iron(III) sulfide and calcium chloride. The balanced equation for this reaction is:

\[2 \text{FeCl}_3 (aq) + 3 \text{CaS} (aq) \rightarrow \text{Fe}_2\text{S}_3 (s) + 3 \text{CaCl}_2 (aq)\]

This means 2 moles of FeCl\(_3\) react with 3 moles of CaS to produce 1 mole of Fe\(_2\)S\(_3\) and 3 moles of CaCl\(_2\). Each coefficient in the equation needs to correctly represent the ratio of molecules involved.

mole concept

The mole concept is a fundamental idea in chemistry. It tells us about the amount of substance. 1 mole of any substance contains Avogadro's number of entities (6.022 × 10\(^23\) particles). In our exercise, we use molarity (moles per liter) and volume to find the moles of each reactant.

For FeCl\(_3\):
\[ \text{Moles of FeCl}_3 = 0.135 \text{ M} \times 62.0 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.00837 \text{ mol} \]

For CaS:
\[ \text{Moles of CaS} = 0.285 \text{ M} \times 45.0 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.012825 \text{ mol} \]

This calculation shows how many moles of each substance we have before the reaction begins.

molar mass calculation

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It incorporates the atomic weights of all atoms in a molecule.

For Fe\(_2\)S\(_3\), calculate it as follows:

• Iron (Fe): Atomic mass = 55.85 g/mol
• Sulfur (S): Atomic mass = 32.06 g/mol

The molar mass of Fe\(_2\)S\(_3\) then:

\[2 \times 55.85 + 3 \times 32.06 = 207.87 + 96.18 = 303.91 \text{ g/mol}\]

This molar mass is necessary to convert moles of Fe\(_2\)S\(_3\) to grams.

stoichiometry

Stoichiometry uses the balanced chemical equation to quantify the reactants and products. In our problem, start by identifying the limiting reactant—it’s the reactant that gets used up first, thus stopping the reaction.

Compare the mole ratio:

For FeCl\(_3\): \[ \frac{0.00837 \text{ mol FeCl}_3}{2} = 0.004185 \text{ mol} \]
For CaS: \[ \frac{0.012825 \text{ mol CaS}}{3} = 0.004275 \text{ mol} \]

0.004185 mol of FeCl\(_3\) limits the reaction. Use this to find moles of Fe\(_2\)S\(_3\) produced:

\[ \text{Moles of Fe}_2\text{S}_3 = \frac{0.00837 \text{ mol FeCl}_3}{2} = 0.004185 \text{ mol} \]

Finally, convert this to grams using molar mass:

\[ \text{Mass of Fe}_2\text{S}_3 = 0.004185 \text{ mol} \times 303.91 \text{ g/mol} = 1.272 \text{ g}\]
So, the reaction produces 1.272 grams of Fe\(_2\)S\(_3\).