Chemistry · General chemistry I · Worked example

Find the limiting reactant from two masses

Aluminum reacts with chlorine: 2Al(s) + 3Cl₂(g) → 2AlCl₃(s). A mixture of 10.0 g Al and 35.0 g Cl₂ produces 38.0 g AlCl₃. Which reactant limits, what is the theoretical yield, and what is the percent yield?

2 Al (s) + 3 {Cl}_{2} (g) \to 2 {AlCl}_{3} (s)

Start from the balanced equation

The coefficients give the mole ratios: 3 mol Cl₂ react with 2 mol Al and make 2 mol AlCl₃. Check the atoms: two Al and six Cl on each side.

2 Al (s) + 3 {Cl}_{2} (g) \to 2 {AlCl}_{3} (s)

Find the molar masses

Use the atomic masses from the periodic table: Al 26.98, Cl 35.45. Chlorine is diatomic, so M(Cl₂) = 70.90 g/mol, and M(AlCl₃) = 26.98 + 3(35.45) = 133.33 g/mol.

M ({AlCl}_{3}) = 26.98 + 3 (35.45) = 133.33 g/mol

\begin{aligned} M ({AlCl}_{3}) & = 26.98 + 3 (35.45) \\ = 133.33 g/mol \end{aligned}

How much product could the chlorine make?

Convert 35.0 g Cl₂ to moles, use the 2 : 3 ratio from the equation, then convert to grams of AlCl₃. The chlorine alone could make 43.9 g.

35.0 g {Cl}_{2} \times \frac{1 mol {Cl}_{2}}{70.90 g {Cl}_{2}} \times \frac{2 mol {AlCl}_{3}}{3 mol {Cl}_{2}} \times \frac{133.33 g {AlCl}_{3}}{1 mol {AlCl}_{3}} = 43.9 g {AlCl}_{3}

\begin{array}{c} 35.0 g {Cl}_{2} \times \frac{1 mol {Cl}_{2}}{70.90 g {Cl}_{2}} \\ \times \frac{2 mol {AlCl}_{3}}{3 mol {Cl}_{2}} \\ \times \frac{133.33 g {AlCl}_{3}}{1 mol {AlCl}_{3}} \\ = 43.9 g {AlCl}_{3} \end{array}

How much could the aluminum make?

Repeat for 10.0 g Al with the 2 : 2 ratio. The aluminum alone could make 49.4 g.

10.0 g Al \times \frac{1 mol Al}{26.98 g Al} \times \frac{2 mol {AlCl}_{3}}{2 mol Al} \times \frac{133.33 g {AlCl}_{3}}{1 mol {AlCl}_{3}} = 49.4 g {AlCl}_{3}

\begin{array}{c} 10.0 g Al \times \frac{1 mol Al}{26.98 g Al} \\ \times \frac{2 mol {AlCl}_{3}}{2 mol Al} \\ \times \frac{133.33 g {AlCl}_{3}}{1 mol {AlCl}_{3}} \\ = 49.4 g {AlCl}_{3} \end{array}

Choose the smaller amount

Chlorine gives less product, so Cl₂ is the limiting reactant and the theoretical yield is 43.9 g AlCl₃. Aluminum is in excess even though there is less aluminum than chlorine by mass.

Calculate the percent yield

Divide the measured 38.0 g by the theoretical 43.9 g and multiply by 100%.

\frac{38.0 g {AlCl}_{3}}{43.9 g {AlCl}_{3}} \times 100 % = 86.6 %

Result

Cl₂ is the limiting reactant. The theoretical yield is 43.9 g AlCl₃, and the percent yield is 86.6%.

Your turn

In the same mixture, how many grams of aluminum are left over when the chlorine runs out?

Show the answer and explanation

1.12 g of aluminum.

35.0 g Cl₂ uses 35.0/70.90 × 2/3 = 0.3291 mol Al, which is 8.88 g. The mixture started with 10.0 g, so 10.0 − 8.88 = 1.12 g of aluminum is left.

\begin{array}{l} 35.0 g {Cl}_{2} \to 8.88 g Al used \\ 10.0 g - 8.88 g = 1.12 g Al \end{array}

Keep exploring

Open the mixture in Stoichiometry & yield and raise the chlorine to 45.0 g: aluminum becomes the limiting reactant.

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