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?

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.

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 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 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\(_3\).