Limiting Reactants and Percent Yield
Determine which reactant limits product formation, calculate theoretical and actual yields, and interpret excess reactants and experimental results.
Limiting Reactants and Percent Yield
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The limiting reactant sets the theoretical yield
A balanced equation gives the mole ratio in which reactants are consumed. The limiting reactant runs out first and therefore determines the greatest possible amount of product. Any other reactant supplied beyond the required ratio is in excess.
percent yield = actual yield ÷ theoretical yield × 100%Theoretical yield comes from stoichiometry. Actual yield is what the experiment produces.
Convert before comparing
When amounts are in grams, convert them to moles before using equation coefficients.
Predict product twice
Calculate the product possible from each reactant; the smaller result identifies the limit.
Excess means leftover
Subtract the amount consumed from the amount initially available.
Theoretical yield is a maximum
It is predicted from the limiting reactant under the model.
Actual yield is measured
Loss, side reactions, incomplete reaction, or purification can lower it.
Use PRODUCT FROM A versus PRODUCT FROM B
1. Balance the equation.
2. Convert each reactant to moles.
3. Independently calculate the product each reactant could form.
4. The smaller product is the theoretical yield; its source is the limiting reactant.
5. Only then calculate excess remaining or percent yield.
Why it works
Choosing the smaller mass or coefficient is unreliable. Comparing predicted product places both reactants on the same basis.
Five forms you should recognize
Problem: For 2H₂ + O₂→2H₂O, 4 mol H₂ reacts with 1 mol O₂.
Four moles H₂ could form 4 mol H₂O. One mole O₂ can form only 2 mol H₂O.
Conclusion: O₂ is limiting and the theoretical yield is 2 mol H₂O.
Problem: In the same reaction, how much H₂ remains?
One mole O₂ consumes 2 mol H₂.
H₂ remaining = 4−2 = 2 molProblem: For 2H₂ + O₂→2H₂O, react 8 g H₂ with 32 g O₂.
8 g H₂÷2 g per mole = 4 mol H₂32 g O₂÷32 g per mole = 1 mol O₂O₂ limits and forms 2 mol H₂O, which has a mass of 2(18) = 36 g.
Problem: Theoretical yield is 22 g CO₂, but 17.6 g is collected.
percent yield = 17.6÷22×100% = 80%Problem: A student reports 108% yield.
Interpretation: The collected material may be wet or impure, or a measurement error occurred. The result does not mean the reaction created matter beyond the theoretical prediction.
Check before you commit
- Selecting the reactant with the smaller mass
- Comparing moles without applying the coefficient ratio
- Calculating yield from the excess reactant
- Using actual yield in place of theoretical yield
- Forgetting to subtract consumed excess reactant
- Accepting a yield above 100% without checking impurities or error
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Limiting Reactants and Percent Yield FAQ
Can the reactant with more moles be limiting?
Yes. A reaction may require several moles of it for every mole of the other reactant.
Why is actual yield often below theoretical yield?
Reactions may be incomplete, side reactions may occur, and product may be lost during transfer or purification.
How is this different from the Stoichiometric Mole Ratios reviewer?
That reviewer practices balanced-equation conversions. This one adds competing reactants, leftovers, and experimental yield.
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