Dr. Ray's Learning Center

Sequential Reactions and Overall Yield

General Chemistry · Limiting Reactants and Percent Yield

Dr. Ray’s Learning Center · covers: Sequential Reactions and Overall Yield

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

Before this lesson:

A bakery turns flour into dough, then dough into loaves. If 90% of the flour ends up in the dough and 80% of the dough becomes good loaves, only 72% of the flour ends up as bread: 80% of 90%, not the 85% halfway between them.

Many products are made the same way, in a sequence of reactions. The product of the first reaction is the starting material of the next. This lesson shows how to carry the stoichiometry through every reaction in the sequence and how to combine the percent yields of the reactions into one overall yield (Figure 1).

  1. Starting material Amass to moles
  2. reaction 1its mole ratio
  3. Intermediate Bin mol
  4. reaction 2its mole ratio
  5. Final product Cmoles to grams

One mole ratio per reaction.Overall yield: multiply the yields as fractions.

Figure 1. Follow the moles from the starting material through the intermediate to the final product, one reaction at a time.

Learning Objectives

After this lesson, you should be able to:

  1. Carry stoichiometry through a sequence of reactions, including the overall yield of a multistep process.

Key Ideas

1. The Product of One Reaction Feeds the Next

In a two-step sequence, a starting material A forms an intermediate B in reaction 1, and B forms the final product C in reaction 2. The intermediate is made in one reaction and used up in the next, so it does not appear in the final answer.

The route from nitrogen to nitric acid is an example. In the Haber process, nitrogen and hydrogen form ammonia. The Ostwald process then turns ammonia into nitric acid; its first step burns the ammonia in oxygen to nitrogen monoxide (Figure 2):

N2+3H2→2NH3\mathrm{N_2} + 3\,\mathrm{H_2} \rightarrow 2\,\mathrm{NH_3}
4NH3+5O2→4NO+6H2O4\,\mathrm{NH_3} + 5\,\mathrm{O_2} \rightarrow 4\,\mathrm{NO} + 6\,\mathrm{H_2O}
From nitrogen to nitrogen monoxideThree boxes from top to bottom. N₂ and H₂, the starting materials, form NH₃ in reaction 1, the Haber process. NH₃ is the intermediate: it is made in reaction 1 and used in reaction 2. With O₂ it forms NO, the final product, and H₂O in reaction 2, the first step of the Ostwald process.From nitrogen to nitrogen monoxideN₂ + H₂starting materialsreaction 1NH₃intermediate: made, then usedreaction 2NO + H₂Ofinal product NOthe intermediate links the two reactions
Figure 2. Ammonia is the intermediate: reaction 1 (the Haber process) makes it and reaction 2 (the first step of the Ostwald process) uses it up.

Key idea: each reaction has its own balanced equation and its own mole ratio.

2. A Chain of Mole Ratios

Convert the mass of the starting material to moles once. Then use the mole ratio of reaction 1 to get the moles of the intermediate, and the mole ratio of reaction 2 to get the moles of the final product. Only at the end do you convert back to grams (Figure 3):

g A→mol A→mol B→mol C→g C\text{g A} \;\rightarrow\; \text{mol A} \;\rightarrow\; \text{mol B} \;\rightarrow\; \text{mol C} \;\rightarrow\; \text{g C}

There is no need to find the mass of the intermediate. Its moles carry straight into the next reaction.

  1. Mass of Ain g
  2. ×1M\times \frac{1}{M}molar mass of A
  3. Moles of Ain mol
  4. ×cBcA\times \frac{c_{\text{B}}}{c_{\text{A}}}reaction 1: B on top
  5. Moles of intermediate Bno need for its mass
  6. ×cCcB\times \frac{c_{\text{C}}}{c_{\text{B}}}reaction 2: C on top
  7. Moles of Cin mol
  8. ×M\times Mmolar mass of C
  9. Mass of Ctheoretical yield
Figure 3. One mole ratio for each reaction, in order. The intermediate is never converted to a mass.

Key idea: one mole ratio per reaction, in order; grams only at the start and at the end.

3. The Overall Yield

Each reaction keeps only a fraction of what the reaction before it made. Write each percent yield as a fraction, multiply the fractions, and multiply by 100%:

overall yield=yield1100%×yield2100%×100%\text{overall yield} = \dfrac{\text{yield}_1}{100\%} \times \dfrac{\text{yield}_2}{100\%} \times 100\%

With a third reaction, multiply by its fraction too. The overall yield is always lower than the yield of every single reaction (Figure 4).

Worked Example 1: what is leftThree bars on one scale, for every 100 g of product that could form. At the start: 100 g. After reaction 1, at a 92.0 percent yield: 92.0 g. After reaction 2, at an 85.0 percent yield of that: 78.20 g, an overall yield of 78.2 percent.Worked Example 1: what is leftCould form100 gAfter reaction 1 (92.0%)92.0 gAfter reaction 2 (85.0%)78.20 geach yield acts on what is left
Figure 4. Each reaction keeps its fraction of what the one before it made: 92.0% of 100 g, then 85.0% of that, leaves 78.20 g.

Key idea: multiply the yields written as fractions; never add or average them.

4. From the Overall Yield to the Actual Yield

The theoretical yield of the final product assumes that every reaction goes to 100%. The overall yield works like any percent yield, as a unit factor that turns the theoretical yield into the actual yield (Figure 5):

mactual=mtheoretical×overall yield100%m_{\text{actual}} = m_{\text{theoretical}} \times \dfrac{\text{overall yield}}{100\%}
Overall yield as a factorA unit factor with 76.50 grams of NO actually obtained on top and 100 grams of theoretical NO on the bottom, from the overall yield of Worked Example 3. Use it when you have a theoretical yield: grams theoretical cancel.Overall yield as a factor76.50 g NO actual100 g NO theoreticaluse when you havea theoretical yield of NOg theoretical cancels
Figure 5. The overall yield of Worked Example 3, 76.50%, written as a unit factor: it turns grams of theoretical NO into grams actually obtained.

Worked Example 1: Overall Yield of Two Reactions

Problem: A product is made by a sequence of 2 reactions. The percent yields of the reactions are 92.0% and 85.0%, in order. What is the overall percent yield?

Step 1. Write each yield as a fraction, multiply the fractions, and multiply by 100%.

overall yield=92.0%100%×85.0%100%×100%=0.920×0.850×100%=78.20%≈78.2%\begin{aligned} \text{overall yield} &= \dfrac{92.0\%}{100\%} \times \dfrac{85.0\%}{100\%} \times 100\% \\ &= 0.920 \times 0.850 \times 100\% \\ &= 78.20\% \approx 78.2\% \end{aligned}

Final Answer: 78.2%

Check: the overall yield is below 85.0%, the lower of the two yields, as it must be.

Your turn: overall yield

Same step as Worked Example 1: multiply the yields written as fractions, then multiply by 100%.

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Worked Example 2: Theoretical Yield through Two Reactions

Problem: Potassium chlorate is heated to give oxygen, and the oxygen is then used to burn magnesium to magnesium oxide. Starting from 15.0 g of potassium chlorate (KClO3), with enough of every other reactant, what mass of MgO can be made in theory (100% yield in both reactions)?

Step 1. Write both balanced equations and the molar masses.

2KClO3→2KCl+3O22\,\mathrm{KClO_3} \rightarrow 2\,\mathrm{KCl} + 3\,\mathrm{O_2}
2Mg+O2→2MgO2\,\mathrm{Mg} + \mathrm{O_2} \rightarrow 2\,\mathrm{MgO}
MKClO3=122.55g/mol and MMgO=40.30g/molM_{\mathrm{KClO_3}} = 122.55\,\text{g/mol} \text{ and } M_{\mathrm{MgO}} = 40.30\,\text{g/mol}

Step 2. Convert the mass of KClO3 to moles.

nKClO3=15.0g122.55g/mol=0.1224moln_{\mathrm{KClO_3}} = \dfrac{15.0\,\text{g}}{122.55\,\text{g/mol}} = 0.1224\,\text{mol}

Step 3. Reaction 1: use its mole ratio to find the moles of O2.

nO2=0.1224mol KClO3×3mol O22mol KClO3=0.1836mol O2\begin{aligned} n_{\mathrm{O_2}} &= 0.1224\,\text{mol}\ \mathrm{KClO_3} \times \dfrac{3\,\text{mol}\ \mathrm{O_2}}{2\,\text{mol}\ \mathrm{KClO_3}} \\ &= 0.1836\,\text{mol}\ \mathrm{O_2} \end{aligned}

Step 4. Reaction 2: use its mole ratio to find the moles of MgO. The O2 made in reaction 1 is the starting material of reaction 2.

nMgO=0.1836mol O2×2mol MgO1mol O2=0.3672mol MgO\begin{aligned} n_{\mathrm{MgO}} &= 0.1836\,\text{mol}\ \mathrm{O_2} \times \dfrac{2\,\text{mol}\ \mathrm{MgO}}{1\,\text{mol}\ \mathrm{O_2}} \\ &= 0.3672\,\text{mol}\ \mathrm{MgO} \end{aligned}

Step 5. Convert the moles of MgO to grams.

mMgO=0.3672mol×40.30g/mol=14.80g≈14.8g\begin{aligned} m_{\mathrm{MgO}} &= 0.3672\,\text{mol} \times 40.30\,\text{g/mol} \\ &= 14.80\,\text{g} \approx 14.8\,\text{g} \end{aligned}

Final Answer: 14.8 g MgO

  1. Potassium chlorate15.0g KClO315.0\,\text{g}\ \mathrm{KClO_3}given
  2. ×1mol122.55g\times \frac{1\,\text{mol}}{122.55\,\text{g}}g to mol
  3. Moles of KClO30.1224mol KClO30.1224\,\text{mol}\ \mathrm{KClO_3}
  4. ×32\times \frac{3}{2}reaction 1: 3 mol O2 per 2 mol KClO3
  5. Moles of O20.1836mol O20.1836\,\text{mol}\ \mathrm{O_2}the intermediate
  6. ×21\times \frac{2}{1}reaction 2: 2 mol MgO per 1 mol O2
  7. Moles of MgO0.3672mol MgO0.3672\,\text{mol}\ \mathrm{MgO}
  8. ×40.30g/mol\times 40.30\,\text{g/mol}mol to g
  9. Magnesium oxide14.8g MgO14.8\,\text{g}\ \mathrm{MgO}Final answer to 3 significant figures
Figure 6. Worked Example 2 one step at a time: moles of KClO3, the mole ratio of each reaction, then grams of MgO.

Keep 4 significant figures in every step (0.1224, 0.1836 and 0.3672 mol) and round only the final answer.

Your turn: theoretical yield through two reactions

Same five steps as Worked Example 2: moles of the starting material, one mole ratio per reaction, then grams of the final product.

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Worked Example 3: Actual Yield of a Two-Step Process

Problem: Nitrogen and hydrogen form ammonia in the Haber process, and the ammonia is then burned in oxygen to nitrogen monoxide, the first step of the Ostwald process. A run starts with 20.0 g of nitrogen (N2) and enough of every other reactant. Reaction 1 has a yield of 85.0% and reaction 2 a yield of 90.0%. What mass of NO is actually obtained?

Step 1. Write both balanced equations and the molar masses.

N2+3H2→2NH3\mathrm{N_2} + 3\,\mathrm{H_2} \rightarrow 2\,\mathrm{NH_3}
4NH3+5O2→4NO+6H2O4\,\mathrm{NH_3} + 5\,\mathrm{O_2} \rightarrow 4\,\mathrm{NO} + 6\,\mathrm{H_2O}
MN2=28.01g/mol and MNO=30.01g/molM_{\mathrm{N_2}} = 28.01\,\text{g/mol} \text{ and } M_{\mathrm{NO}} = 30.01\,\text{g/mol}

Step 2. Convert the mass of N2 to moles, then apply the mole ratio of each reaction.

nN2=20.0g28.01g/mol=0.7140moln_{\mathrm{N_2}} = \dfrac{20.0\,\text{g}}{28.01\,\text{g/mol}} = 0.7140\,\text{mol}
nNH3=0.7140mol N2×2mol NH31mol N2=1.428mol NH3\begin{aligned} n_{\mathrm{NH_3}} &= 0.7140\,\text{mol}\ \mathrm{N_2} \times \dfrac{2\,\text{mol}\ \mathrm{NH_3}}{1\,\text{mol}\ \mathrm{N_2}} \\ &= 1.428\,\text{mol}\ \mathrm{NH_3} \end{aligned}
nNO=1.428mol NH3×4mol NO4mol NH3=1.428mol NO\begin{aligned} n_{\mathrm{NO}} &= 1.428\,\text{mol}\ \mathrm{NH_3} \times \dfrac{4\,\text{mol}\ \mathrm{NO}}{4\,\text{mol}\ \mathrm{NH_3}} \\ &= 1.428\,\text{mol}\ \mathrm{NO} \end{aligned}

Step 3. Theoretical yield: convert the moles of NO to grams.

mNO,theoretical=1.428mol×30.01g/mol=42.85g\begin{aligned} m_{\mathrm{NO},\,\text{theoretical}} &= 1.428\,\text{mol} \times 30.01\,\text{g/mol} \\ &= 42.85\,\text{g} \end{aligned}

Step 4. Find the overall yield of the two reactions.

overall yield=85.0%100%×90.0%100%×100%=0.850×0.900×100%=76.50%\begin{aligned} \text{overall yield} &= \dfrac{85.0\%}{100\%} \times \dfrac{90.0\%}{100\%} \times 100\% \\ &= 0.850 \times 0.900 \times 100\% \\ &= 76.50\% \end{aligned}

Step 5. Multiply the theoretical yield by the overall yield written as a unit factor. g theoretical cancels:

mNO,actual=42.85g theoretical×76.50g actual100g theoretical=32.78g≈32.8g\begin{aligned} m_{\mathrm{NO},\,\text{actual}} &= 42.85\,\text{g theoretical} \\ &\;\; \times \dfrac{76.50\,\text{g actual}}{100\,\text{g theoretical}} \\ &= 32.78\,\text{g} \approx 32.8\,\text{g} \end{aligned}

Final Answer: 32.8 g NO

  1. Nitrogen20.0g N220.0\,\text{g}\ \mathrm{N_2}given
  2. ×1mol28.01g\times \frac{1\,\text{mol}}{28.01\,\text{g}}g to mol
  3. Moles of N20.7140mol N20.7140\,\text{mol}\ \mathrm{N_2}
  4. ×21\times \frac{2}{1}reaction 1: 2 mol NH3 per 1 mol N2
  5. Moles of NH31.428mol NH31.428\,\text{mol}\ \mathrm{NH_3}the intermediate
  6. ×44\times \frac{4}{4}reaction 2: 4 mol NO per 4 mol NH3
  7. Moles of NO1.428mol NO1.428\,\text{mol}\ \mathrm{NO}
  8. ×30.01g/mol\times 30.01\,\text{g/mol}mol to g
  9. Theoretical yield42.85g NO42.85\,\text{g}\ \mathrm{NO}4 significant figures
  10. ×76.50100\times \frac{76.50}{100}overall yield 85.0% of 90.0%
  11. Nitrogen monoxide32.8g NO32.8\,\text{g}\ \mathrm{NO}Final answer to 3 significant figures
Figure 7. Worked Example 3 one step at a time: the theoretical yield through both reactions, then the overall yield.

Check: 32.8 g is less than the theoretical 42.85 g, and the overall yield (76.50%) is below both reaction yields.

Your turn: actual yield of a two-step process

Same five steps as Worked Example 3: the theoretical yield through both reactions, the overall yield, then the actual yield.

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Common Mistakes

Common mistake: Averaging the yields. In Worked Example 1, the mean of 92.0% and 85.0% is 88.5%, too high. Each reaction keeps only its fraction of what the one before it made, so the overall yield is 78.2%.

Common mistake: Using only the lowest yield. 85.0% is also too high: the first reaction already lost 8.0% of the product. Multiply every yield.

Common mistake: Stopping at the intermediate. In Worked Example 2, 0.1836 mol of O2 is 5.88 g of oxygen, not the MgO that was asked. Carry the moles through reaction 2.

Common mistake: Leaving out one mole ratio. Skipping the 2 mol MgO per 1 mol O2 of reaction 2 gives 7.40 g, half the right answer. Use one mole ratio for each reaction.

Common mistake: Applying only one yield. In Worked Example 3, using only the 85.0% of reaction 1 gives 36.4 g; averaging the two yields gives 37.5 g. Both are too high: the overall yield is 76.50%.

Exam tip: the overall yield is lower than every single yield, and the actual yield is lower than the theoretical yield. A larger answer means a step was skipped.

Key Takeaway

In a sequence of reactions, the product of one reaction is the starting material of the next. Convert the starting mass to moles, apply one mole ratio per reaction, and convert the moles of the final product to grams. The overall yield is the product of the reaction yields written as fractions, times 100%; as a unit factor, it turns the theoretical yield of the final product into the actual yield.

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 reaction sequences and their contexts are original teaching examples. The practice problems are generated and checked by an audited engine; see the lesson audit log.