Dr. Ray's Learning Center

Limiting Reactant and Theoretical Yield

General Chemistry · Limiting Reactants and Percent Yield

Dr. Ray’s Learning Center · covers: Limiting Reactant and Theoretical Yield

  • 20–25 min
  • 8 figures
  • 3 worked examples
  • 4 practice sets

Before this lesson:

A bicycle shop has 6 frames and 10 wheels. Each bicycle needs 1 frame and 2 wheels, so the shop can build only 5 bicycles: the wheels run out first, and 1 frame is left over. The wheels limit the number of bicycles, even though there are more wheels than frames.

Reactions work the same way. Reactants are rarely mixed in exactly the ratio of the balanced equation, so one of them runs out first and stops the reaction. This lesson shows how to find that reactant and the largest amount of product it allows (Figure 1).

  1. Both reactantsmasses or moles
  2. ×1M\times \frac{1}{M}masses to moles
  3. Moles of eachnot yet comparable
  4. ×cPcA\times \frac{c_{\text{P}}}{c_{\text{A}}}product from each
  5. Smaller amount of productlimiting reactant and theoretical yield

Compare the product, not the reactants.The other reactant is in excess.

Figure 1. Find how much product each reactant could form; the smaller amount names the limiting reactant and is the theoretical yield.

Learning Objectives

After this lesson, you should be able to:

  1. Identify the limiting reactant from given masses or amounts.
  2. Calculate the theoretical yield from the limiting reactant.

Key Ideas

1. The Limiting Reactant

The limiting reactant is the reactant that is used up first; when it is gone, the reaction stops. Every other reactant is in excess: some of it is left over (Figure 2).

The wheels limit the number of bicyclesSix frames and ten wheels. Each bicycle needs one frame and two wheels, so five bicycles can be built; all ten wheels are used and one frame is left over. The wheels are limiting even though there are more wheels than frames.Building bicycleshave: 6 frames, 10 wheelseach bicycle: 1 frame and 2 wheels5 bicycles: the wheels run out1 frame left over (in excess)
Figure 2. The part that runs out first limits how many bicycles can be built. Wheels are limiting here although there are more wheels than frames, because each bicycle needs two of them.

The theoretical yield is the amount of product formed when all of the limiting reactant reacts. It is the most product the reaction mixture can give.

Key idea: the limiting reactant, not the reactant you have most of, decides how much product forms.

2. Compare the Product Each Reactant Can Form

To find the limiting reactant, convert each reactant to moles, then use the mole ratio to find how much product each one could form if it reacted completely:

nP, from A=nA×cPcAn_{\text{P, from A}} = n_{\text{A}} \times \dfrac{c_{\text{P}}}{c_{\text{A}}}

Here cAc_{\text{A}} and cPc_{\text{P}} are the coefficients of reactant A and product P in the balanced equation. Do this for each reactant. The reactant that forms less product is limiting, and that smaller amount is the theoretical yield (Figure 3).

  1. Reactant Amoles of A
  2. ×cPcA\times \frac{c_{\text{P}}}{c_{\text{A}}}product A could form
  3. Reactant Bmoles of B
  4. ×cPcB\times \frac{c_{\text{P}}}{c_{\text{B}}}product B could form
  5. Comparethe smaller amount wins
Figure 3. One calculation for each reactant, then one comparison. The reactant that could form less product runs out first.

Key idea: the smaller amount of product wins: it names the limiting reactant and gives the theoretical yield.

3. The Smaller Mass Is Not Always Limiting

Neither the mass nor the number of moles alone decides. Masses must first become moles, and moles must be compared through the coefficients. In Figure 4, there are more H2 molecules than O2 molecules, yet H2 runs out first, because each O2 needs two H2:

2H2+O2→2H2O2\,\mathrm{H_2} + \mathrm{O_2} \rightarrow 2\,\mathrm{H_2O}
More hydrogen molecules, yet hydrogen is limitingBefore the reaction: four hydrogen molecules and three oxygen molecules. After: four water molecules and one oxygen molecule left over. Each oxygen molecule needs two hydrogen molecules, so the hydrogen runs out first even though there were more hydrogen molecules.Before: 4 H₂ and 3 O₂HHHHHHHHOOOOOOAfter: 4 H₂O and 1 O₂ leftOHHOHHOHHOHHOOH₂ is used up first: it is limiting
Figure 4. Two H2 for every O2: 4 H2 react with only 2 of the 3 O2. H2 is limiting although there are more H2 molecules.

Key idea: compare the amounts of product, not the masses or the moles of the reactants.

4. From Moles of Product to Theoretical Yield

The molar mass of the product converts the theoretical yield from moles to a mass, as in mass-to-mass stoichiometry (Figure 5). In the laboratory the actual yield is usually smaller; the next lessons compare the two and find how much of the excess reactant is left.

  1. Smaller amount of productin mol, from the limiting reactant
  2. ×M\times Mmolar mass of the product
  3. Theoretical yieldin g, the most product possible

Actual yield: the mass you weigh.It is usually smaller than the theoretical yield.

Figure 5. Only the smaller amount of product is converted to grams; the other amount is never reached because the limiting reactant runs out.

Worked Example 1: Limiting Reactant from Mole Amounts

Problem: Nitrogen reacts with hydrogen to produce ammonia. A reaction mixture contains 2.00 mol of N2 and 4.50 mol of H2. Which reactant is limiting, and what is the theoretical yield of NH3 in mol?

Step 1. Write the balanced equation.

N2+3H2→2NH3\mathrm{N_2} + 3\,\mathrm{H_2} \rightarrow 2\,\mathrm{NH_3}

Step 2. Find the moles of NH3 each reactant could form.

nNH3,from N2=2.00mol N2×2mol NH31mol N2=4.000mol NH3\begin{aligned} n_{\mathrm{NH_3},\,\text{from}\ \mathrm{N_2}} &= 2.00\,\text{mol}\ \mathrm{N_2} \times \dfrac{2\,\text{mol}\ \mathrm{NH_3}}{1\,\text{mol}\ \mathrm{N_2}} \\ &= 4.000\,\text{mol}\ \mathrm{NH_3} \end{aligned}
nNH3,from H2=4.50mol H2×2mol NH33mol H2=3.000mol NH3\begin{aligned} n_{\mathrm{NH_3},\,\text{from}\ \mathrm{H_2}} &= 4.50\,\text{mol}\ \mathrm{H_2} \times \dfrac{2\,\text{mol}\ \mathrm{NH_3}}{3\,\text{mol}\ \mathrm{H_2}} \\ &= 3.000\,\text{mol}\ \mathrm{NH_3} \end{aligned}

Step 3. Compare. H2 forms less NH3, so H2 is the limiting reactant and N2 is in excess. The theoretical yield is the smaller amount:

nNH3=3.000mol≈3.00moln_{\mathrm{NH_3}} = 3.000\,\text{mol} \approx 3.00\,\text{mol}

Final Answer: H2 is limiting; theoretical yield 3.00 mol NH3

Worked Example 1: NH₃ from eachBars for the moles of ammonia each reactant could form: 4.000 mol from 2.00 mol of nitrogen and 3.000 mol from 4.50 mol of hydrogen. The hydrogen bar is shorter and marked: hydrogen is limiting and the theoretical yield is 3.00 mol.Worked Example 1: NH₃ from eachfrom 2.00 mol N₂ and 4.50 mol H₂from N₂: 4.000 molfrom H₂: 3.000 molshorter bar: H₂ is limiting
Figure 6. The reactant with the shorter bar runs out first. H2 could form only 3.000 mol of NH3, so that is the theoretical yield.

Check: all 2.00 mol of N2 would need 6.00 mol of H2, and only 4.50 mol is there. H2 runs out first.

Your turn: limiting reactant from mole amounts

Same three steps as Worked Example 1: the equation, the moles of product from each reactant, then the smaller amount.

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Worked Example 2: Identifying the Limiting Reactant from Masses

Problem: Magnesium burns in oxygen to form magnesium oxide. A mixture contains 10.0 g of Mg and 10.0 g of O2. Which reactant is limiting?

Step 1. Write the balanced equation and the molar masses.

2Mg+O2→2MgO2\,\mathrm{Mg} + \mathrm{O_2} \rightarrow 2\,\mathrm{MgO}
MMg=24.31g/mol and MO2=32.00g/molM_{\mathrm{Mg}} = 24.31\,\text{g/mol} \text{ and } M_{\mathrm{O_2}} = 32.00\,\text{g/mol}

Step 2. Convert each mass to moles.

nMg=10.0g24.31g/mol=0.4114moln_{\mathrm{Mg}} = \dfrac{10.0\,\text{g}}{24.31\,\text{g/mol}} = 0.4114\,\text{mol}
nO2=10.0g32.00g/mol=0.3125moln_{\mathrm{O_2}} = \dfrac{10.0\,\text{g}}{32.00\,\text{g/mol}} = 0.3125\,\text{mol}

Step 3. Find the moles of MgO each reactant could form.

nMgO,from Mg=0.4114mol Mg×2mol MgO2mol Mg=0.4114mol MgO\begin{aligned} n_{\mathrm{MgO},\,\text{from}\ \mathrm{Mg}} &= 0.4114\,\text{mol}\ \mathrm{Mg} \times \dfrac{2\,\text{mol}\ \mathrm{MgO}}{2\,\text{mol}\ \mathrm{Mg}} \\ &= 0.4114\,\text{mol}\ \mathrm{MgO} \end{aligned}
nMgO,from O2=0.3125mol O2×2mol MgO1mol O2=0.6250mol MgO\begin{aligned} n_{\mathrm{MgO},\,\text{from}\ \mathrm{O_2}} &= 0.3125\,\text{mol}\ \mathrm{O_2} \times \dfrac{2\,\text{mol}\ \mathrm{MgO}}{1\,\text{mol}\ \mathrm{O_2}} \\ &= 0.6250\,\text{mol}\ \mathrm{MgO} \end{aligned}

Step 4. Compare. Mg forms less MgO, so Mg is limiting and O2 is in excess.

Final Answer: magnesium (Mg)

Worked Example 2: Mg and O₂Three pairs of bars for magnesium and oxygen. Mass: 10.0 grams each, equal. Moles: 0.4114 for magnesium and 0.3125 for oxygen, so oxygen has fewer moles. Magnesium oxide each could form: 0.4114 mol from magnesium and 0.6250 mol from oxygen. Only this last comparison decides: magnesium forms less and is limiting.Worked Example 2: Mg and O₂massMg: 10.0 gO₂: 10.0 gmolesMg: 0.4114 molO₂: 0.3125 molMgO each could formfrom Mg: 0.4114 molfrom O₂: 0.6250 molonly the last pair decides: Mg is limiting
Figure 7. Equal masses, fewer moles of O2, and still Mg is limiting: only the amounts of product decide.

The masses are equal, and there are fewer moles of O2, yet Mg is limiting: each O2 reacts with two Mg atoms. Only the amounts of product settle the question.

Your turn: identifying the limiting reactant

Same four steps as Worked Example 2: molar masses, moles of each reactant, moles of product from each, then compare.

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Worked Example 3: Theoretical Yield as a Mass

Problem: Aluminum reacts with chlorine gas to form aluminum chloride. A reactor is charged with 12.0 g of Al and 35.0 g of Cl2. What is the theoretical yield of AlCl3 in g?

Step 1. Write the balanced equation and the molar masses.

2Al+3Cl2→2AlCl32\,\mathrm{Al} + 3\,\mathrm{Cl_2} \rightarrow 2\,\mathrm{AlCl_3}
MAl=26.98g/molMCl2=70.90g/molMAlCl3=133.33g/mol\begin{aligned} M_{\mathrm{Al}} &= 26.98\,\text{g/mol} \\ M_{\mathrm{Cl_2}} &= 70.90\,\text{g/mol} \\ M_{\mathrm{AlCl_3}} &= 133.33\,\text{g/mol} \end{aligned}

Step 2. Convert each mass to moles.

nAl=12.0g26.98g/mol=0.4448moln_{\mathrm{Al}} = \dfrac{12.0\,\text{g}}{26.98\,\text{g/mol}} = 0.4448\,\text{mol}
nCl2=35.0g70.90g/mol=0.4937moln_{\mathrm{Cl_2}} = \dfrac{35.0\,\text{g}}{70.90\,\text{g/mol}} = 0.4937\,\text{mol}

Step 3. Find the moles of AlCl3 each reactant could form.

nAlCl3,from Al=0.4448mol Al×2mol AlCl32mol Al=0.4448mol AlCl3\begin{aligned} n_{\mathrm{AlCl_3},\,\text{from}\ \mathrm{Al}} &= 0.4448\,\text{mol}\ \mathrm{Al} \times \dfrac{2\,\text{mol}\ \mathrm{AlCl_3}}{2\,\text{mol}\ \mathrm{Al}} \\ &= 0.4448\,\text{mol}\ \mathrm{AlCl_3} \end{aligned}
nAlCl3,from Cl2=0.4937mol Cl2×2mol AlCl33mol Cl2=0.3291mol AlCl3\begin{aligned} n_{\mathrm{AlCl_3},\,\text{from}\ \mathrm{Cl_2}} &= 0.4937\,\text{mol}\ \mathrm{Cl_2} \times \dfrac{2\,\text{mol}\ \mathrm{AlCl_3}}{3\,\text{mol}\ \mathrm{Cl_2}} \\ &= 0.3291\,\text{mol}\ \mathrm{AlCl_3} \end{aligned}

Cl2 forms less AlCl3, so Cl2 is the limiting reactant and Al is in excess.

Step 4. Convert the theoretical moles of AlCl3 to grams.

mAlCl3=0.3291mol×133.33g/mol=43.88g≈43.9g\begin{aligned} m_{\mathrm{AlCl_3}} &= 0.3291\,\text{mol} \times 133.33\,\text{g/mol} \\ &= 43.88\,\text{g} \approx 43.9\,\text{g} \end{aligned}

Final Answer: 43.9 g AlCl3

  1. Reactants12.0g Al and 35.0g Cl212.0\,\text{g}\ \mathrm{Al} \text{ and } 35.0\,\text{g}\ \mathrm{Cl_2}given masses
  2. ×1M\times \frac{1}{M}g to mol
  3. Moles0.4448mol Al and 0.4937mol Cl20.4448\,\text{mol}\ \mathrm{Al} \text{ and } 0.4937\,\text{mol}\ \mathrm{Cl_2}
  4. ×cPcA\times \frac{c_{\text{P}}}{c_{\text{A}}}2 per 2 and 2 per 3
  5. AlCl3 each could form0.4448mol and 0.3291mol AlCl30.4448\,\text{mol} \text{ and } 0.3291\,\text{mol}\ \mathrm{AlCl_3}from Al and from Cl2
  6. ×133.33g/mol\times 133.33\,\text{g/mol}the smaller amount, mol to g
  7. Theoretical yield43.9g AlCl343.9\,\text{g}\ \mathrm{AlCl_3}Final answer to 3 significant figures
Figure 8. Worked Example 3 at a glance: moles of each reactant, the AlCl3 each could form, and the smaller amount converted to grams.

Cl2 has the larger mass and even more moles than Al, but it is still limiting: each 2 Al need 3 Cl2.

Your turn: theoretical yield as a mass

Same four steps as Worked Example 3, with other reactions and masses.

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Common Mistakes

Common mistake: Choosing the reactant with the smaller mass. In Worked Example 3, Al has the smaller mass (12.0 g against 35.0 g), but Cl2 is limiting. Convert masses to moles of product before you compare.

Common mistake: Comparing moles without the coefficients. In Worked Example 2 there are fewer moles of O2 (0.3125 against 0.4114), but Mg is limiting. Use the mole ratio to find the product each reactant can form.

Common mistake: Using the excess reactant. Starting from Al in Worked Example 3 gives 59.3 g of AlCl3, more than the 43.9 g the chlorine allows. The theoretical yield is always the smaller amount.

Common mistake: Stopping at moles. When the yield is asked as a mass, finish with the molar mass of the product: 0.3291 mol of AlCl3 is 43.9 g.

Key Takeaway

The limiting reactant runs out first and sets the theoretical yield; the other reactants are in excess. Convert each reactant to moles, use the mole ratio to find the product each could form, and take the smaller amount. Neither the smaller mass nor the smaller number of moles is a safe shortcut.

The relations to keep at hand:

nP, from A=nA×cPcAn_{\text{P, from A}} = n_{\text{A}} \times \dfrac{c_{\text{P}}}{c_{\text{A}}}

Practice: randomized problems

Each problem below uses fresh numbers every time: work it out, check your answer, then open the worked solution to compare your steps.

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Your results

This summary collects every practice problem you try on this page and suggests what to review next.

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Figures: original diagrams by Dr. Ray’s Learning Center, licensed CC BY 4.0. Atomic weights: IUPAC Commission on Isotopic Abundances and Atomic Weights, standard atomic weights 2021 (abridged values), the same table as the other stoichiometry lessons. The practice problems are generated and checked by an audited engine; see the lesson audit log.