Limiting Reactant and Yield
Limiting Reactant
In experimental chemistry, one or more reactants are often used in excess so that the other, more expensive or important, reactant is completely consumed. The reactant that is completely used up first and thus controls the amount of product formed is called the Limiting Reactant. Once it is consumed, the reaction stops and no additional product can form. The reactant that remains unreacted is called the Excess Reactant.
Purpose of excess: An excess of one reactant ensures the other reactant is fully consumed, avoiding wastage of expensive materials.
Reaction speed: Sometimes an excess reactant (such as oxygen) is used deliberately to make the reaction proceed faster.
Product limitation: The quantity of every product in the reaction is determined by the limiting reactant — no amount of excess reactant can produce more product once the limiting reactant is gone.
To identify the Limiting Reactant in a reaction where the quantities of two or more reactants are given, follow three systematic steps using the Stoichiometry of the balanced equation.
Step 1 — Convert to moles: Calculate the number of moles of each reactant from the given mass using $n = \frac{m}{M}$ where $m$ is the given mass and $M$ is the molar mass.
Step 2 — Predict product moles: Using the mole ratio from the balanced chemical equation, calculate how many moles of product each reactant can produce.
Step 3 — Identify limiting reactant: The reactant that produces the smaller (least) number of moles of product is the limiting reactant.
Once the Limiting Reactant is identified, you can calculate how much of the Excess Reactant remains unreacted by using the stoichiometric ratio from the balanced equation.
Find moles needed: From the balanced equation, determine how many moles of the excess reactant are required to fully react with all of the limiting reactant.
Subtract from initial: The difference between the initial moles of the excess reactant and the moles that actually reacted gives the moles left over.
Convert back to mass: Multiply the leftover moles by the molar mass of the excess reactant to get the mass remaining unreacted.
Yield — Theoretical, Actual and Percentage
The amount of product calculated from a balanced chemical equation, assuming perfect conditions, is called the Theoretical Yield. It represents the maximum possible amount of product. The amount of product actually obtained from a real experiment is called the Actual Yield. The ratio of actual yield to theoretical yield, expressed as a percentage, is the Percentage Yield, which measures the efficiency of the reaction.
$$\%\text{ yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$$
Percentage yield expresses how efficient a chemical reaction is by comparing the product actually obtained to the maximum possible product from stoichiometry.
$\text{Actual yield}$=The mass of product obtained from the experiment(g or kg)
$\text{Theoretical yield}$=The maximum mass of product predicted by stoichiometric calculation(g or kg)
$\%\text{ yield}$=Efficiency of the reaction as a percentage(%)
$\text{Actual yield} = \text{Theoretical yield}$
→Percentage yield = 100% (ideal, rarely achieved)
$\text{Actual yield} < \text{Theoretical yield}$
→Percentage yield < 100% (the typical case)
Theoretical yield: Always calculated from the limiting reactant using the balanced equation — it assumes no losses.
Actual yield: Always less than theoretical yield in practice due to experimental imperfections.
Percentage yield: A value between 0% and 100% that tells you how efficient the reaction was. Higher percentage means less wastage.
Reasons Actual Yield is Less Than Theoretical Yield
•
Incomplete reactions — some reactant may not fully convert to product
•
Side reactions — reactants may form undesired by-products
•
Mechanical losses during workup — filtration, washing, transfer, and crystallization can lose material
•
Impure reactants — starting materials that are not 100% pure give less product
•
Practical handling errors — inexperienced technique by the operator
Application Strategy
Worked Examples
Problem Statement
Consider the reaction $2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$. If 2 moles of $H_2$ (4 g) react with 2 moles of $O_2$ (64 g), identify the limiting reactant and determine the mass of water produced.
limiting reactantstoichiometry
Strategic Toolkit
$$2H_2 + O_2 \rightarrow 2H_2O$$
Step 1
Compare moles of each reactant to product
From the balanced equation, 2 moles of H₂ produce 2 moles of H₂O (1:1 ratio), and 1 mole of O₂ produces 2 moles of H₂O (1:2 ratio). So 2 moles of H₂ can produce 2 moles of H₂O, and 2 moles of O₂ can produce 4 moles of H₂O.
$$n_{H_2O} = 2 \text{ mol (from } H_2\text{)} \quad \text{vs} \quad n_{H_2O} = 4 \text{ mol (from } O_2\text{)}$$
Step 2
Identify the limiting reactant
H₂ produces fewer moles of product (2 mol vs 4 mol), so hydrogen is the limiting reactant. Oxygen is in excess.
$$\text{Limiting reactant} = H_2$$
Step 3
Calculate mass of water produced
2 moles of H₂O are produced. Molar mass of H₂O = 18 g mol⁻¹.
$$m_{H_2O} = 2 \text{ mol} \times 18 \text{ g mol}^{-1} = 36 \text{ g}$$
Problem Statement
$NH_3$ gas is prepared by heating $NH_4Cl$ and $Ca(OH)_2$: $2NH_4Cl(s) + Ca(OH)_2(s) \rightarrow CaCl_2(s) + 2NH_3(g) + 2H_2O(l)$. If 100 g of each solid is heated, calculate (a) the mass of $NH_3$ produced and (b) the mass of excess reactant left unreacted.
limiting reactantexcess reactantstoichiometry
Strategic Toolkit
$$2NH_4Cl + Ca(OH)_2 \rightarrow CaCl_2 + 2NH_3 + 2H_2O$$
Step 1
Convert masses to moles
Molar mass of NH₄Cl = 53.5 g mol⁻¹, molar mass of Ca(OH)₂ = 74 g mol⁻¹.
$$n_{NH_4Cl} = \frac{100}{53.5} = 1.87 \text{ mol}, \quad n_{Ca(OH)_2} = \frac{100}{74} = 1.35 \text{ mol}$$
Step 2
Find moles of NH₃ each reactant can produce
From the equation, 2 mol NH₄Cl → 2 mol NH₃ (1:1), and 1 mol Ca(OH)₂ → 2 mol NH₃ (1:2).
$$\text{From } NH_4Cl: 1.87 \text{ mol } NH_3, \quad \text{From } Ca(OH)_2: 1.35 \times 2 = 2.70 \text{ mol } NH_3$$
Step 3
Identify limiting reactant and find product mass
NH₄Cl produces less NH₃, so it is the limiting reactant. Mass of NH₃ = 1.87 × 17 g mol⁻¹.
$$m_{NH_3} = 1.87 \times 17 = 31.79 \text{ g}$$
Step 4
Calculate excess reactant remaining
From the equation, 2 mol NH₄Cl react with 1 mol Ca(OH)₂, so 1.87 mol NH₄Cl requires 0.935 mol Ca(OH)₂. Ca(OH)₂ taken = 1.35 mol, so excess = 1.35 − 0.935 = 0.415 mol.
$$m_{excess} = 0.415 \times 74 = 30.71 \text{ g}$$
Problem Statement
When limestone ($CaCO_3$) is roasted, quicklime ($CaO$) is produced: $CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$. The actual yield of $CaO$ is 2.5 kg when 4.5 kg of limestone is roasted. Calculate the percentage yield.
theoretical yieldactual yieldpercentage yield
Strategic Toolkit
$$\%\text{ yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$$
Step 1
Convert mass to consistent units
Convert kg to grams: 4.5 kg = 4500 g of CaCO₃, actual yield = 2.5 kg = 2500 g of CaO.
$$m_{CaCO_3} = 4500 \text{ g}, \quad \text{Actual yield} = 2500 \text{ g}$$
Step 2
Calculate theoretical yield
Molar mass of CaCO₃ = 100 g mol⁻¹, molar mass of CaO = 56 g mol⁻¹. From the equation, 100 g of CaCO₃ produces 56 g of CaO.
$$\text{Theoretical yield} = \frac{56}{100} \times 4500 = 2520 \text{ g}$$
Step 3
Calculate percentage yield
Divide actual yield by theoretical yield and multiply by 100.
$$\%\text{ yield} = \frac{2500}{2520} \times 100 = 99.2\%$$
Problem Statement
Silicon carbide (SiC) is produced by $SiO_2 + 3C \rightarrow SiC + 2CO$. When 100 kg of sand ($SiO_2$) is reacted with excess carbon, the percentage yield is 77%. Calculate the actual yield of SiC in kg. (Molar mass of $SiO_2$ = 60 g mol⁻¹, SiC = 40 g mol⁻¹)
theoretical yieldactual yieldpercentage yield
Strategic Toolkit
$$\%\text{ yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$$
Step 1
Calculate theoretical yield in kg
100 kg of SiO₂. Molar mass ratio: 60 g SiO₂ → 40 g SiC. So 100 kg SiO₂ → (40/60) × 100 kg SiC.
$$\text{Theoretical yield} = \frac{40}{60} \times 100 = 66.67 \text{ kg}$$
Step 2
Calculate actual yield from percentage yield
Rearrange the percentage yield formula: Actual yield = (% yield / 100) × Theoretical yield.
$$\text{Actual yield} = \frac{77}{100} \times 66.67 = 51.4 \text{ kg}$$
Problem Statement
Aluminium sulphide is produced by $2Al + 3S \rightarrow Al_2S_3$. Calculate the mass of $Al_2S_3$ produced when 20 g of Al reacts with 30 g of sulphur, and find the mass of the excess reactant left over. (Molar mass: Al = 27, S = 32, Al₂S₃ = 150 g mol⁻¹)
limiting reactantexcess reactantstoichiometry
Strategic Toolkit
$$2Al + 3S \rightarrow Al_2S_3$$
Step 1
Convert masses to moles
Moles of Al = 20/27 = 0.741 mol, Moles of S = 30/32 = 0.9375 mol.
$$n_{Al} = 0.741 \text{ mol}, \quad n_S = 0.9375 \text{ mol}$$
Step 2
Identify limiting reactant using shortcut
Divide moles by coefficients: Al → 0.741/2 = 0.370, S → 0.9375/3 = 0.3125. Sulphur gives the smaller value, so S is the limiting reactant.
$$\frac{n_{Al}}{2} = 0.370 > \frac{n_S}{3} = 0.3125 \implies \text{S is limiting}$$
Step 3
Calculate mass of Al₂S₃ produced
From the equation, 3 mol S → 1 mol Al₂S₃. So 0.9375 mol S produces 0.9375/3 = 0.3125 mol Al₂S₃.
$$m_{Al_2S_3} = 0.3125 \times 150 = 46.87 \text{ g}$$
Step 4
Calculate excess Al remaining
From the equation, 1 mol Al₂S₃ needs 2 mol Al. So 0.3125 mol Al₂S₃ needs 0.625 mol Al. Al taken = 0.741 mol. Excess = 0.741 − 0.625 = 0.116 mol.
$$m_{excess\ Al} = 0.116 \times 27 = 3.13 \text{ g}$$
Memorize formulas
Percentage Yield
Use to calculate reaction efficiency. Actual yield and theoretical yield must be in the same units. Result is always ≤ 100%.
Formula
$$\%\text{ yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$$
Moles from Mass
Use to convert given mass of a substance to moles. M is molar mass in g mol⁻¹, m is mass in grams.
Formula
$$n = \frac{m}{M}$$
Mass from Moles
Use to convert moles back to mass. Multiply moles by molar mass.
Formula
$$m = n \times M$$
Review
1
Which statement best describes the limiting reactant in a chemical reaction?
Ans
The reactant that is completely consumed first, limiting the amount of product formed
Review
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Definition of Limiting Reactant
The limiting reactant is the substance that is completely used up first in a reaction and determines the theoretical yield of product.
Limiting reactant = fully consumed, controls product amount
2
In the reaction $2H_2 + O_2 \rightarrow 2H_2O$, if 4 mol of $H_2$ are mixed with 1 mol of $O_2$, what role does $H_2$ play?
Ans
Excess reactant
Review
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Check Stoichiometric Ratio
4 mol $H_2$ requires $4 \div 2 \times 1 = 2$ mol $O_2$
$$O_2 \text{ needed} = 2 \text{ mol}$$
Compare with Available
Only 1 mol $O_2$ is available, so $O_2$ is limiting and $H_2$ is in excess.
$$H_2 \text{ is the excess reactant}$$
3
Which of the following correctly defines theoretical yield?
Ans
The maximum amount of product predicted by stoichiometric calculations
Review
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Theoretical Yield Definition
It is the maximum product amount calculated from stoichiometry, assuming perfect reaction conditions.
$$\text{Theoretical yield} = \text{maximum possible product (100\% conversion)}$$
4
For the reaction $C + O_2 \rightarrow CO_2$, 12 g of carbon should theoretically produce 44 g of $CO_2$. If only 33 g of $CO_2$ is collected, what is the percentage yield?
Ans
75%
Review
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Apply Percentage Yield Formula
Divide the actual yield by the theoretical yield and multiply by 100.
$$\%\text{ yield} = \frac{33}{44} \times 100$$
Calculate
$33 \div 44 = 0.75$
$$\%\text{ yield} = 75\%$$
5
A reaction has a theoretical yield of 25 g and an actual yield of 20 g. What is the percentage yield?
Ans
80%
Review
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Apply Formula
Percentage yield = (Actual yield / Theoretical yield) × 100
$$\%\text{ yield} = \frac{20}{25} \times 100$$
Calculate
$20 \div 25 = 0.80$
$$\%\text{ yield} = 80\%$$
6
$N_2 + 3H_2 \rightarrow 2NH_3$. If 28 g of $N_2$ and 12 g of $H_2$ are allowed to react completely, which is the limiting reactant?
Ans
$N_2$
Review
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Calculate Moles of Each Reactant
$n_{N_2} = 28/28 = 1$ mol, $n_{H_2} = 12/2 = 6$ mol
$$n_{N_2} = 1 \text{ mol}, \quad n_{H_2} = 6 \text{ mol}$$
Determine Which Runs Out First
1 mol $N_2$ needs 3 mol $H_2$ (available: 6 mol). 6 mol $H_2$ needs 2 mol $N_2$ (available: 1 mol). $N_2$ is insufficient.
$$N_2 \text{ is the limiting reactant}$$
7
$2Mg + O_2 \rightarrow 2MgO$. If 12 g of $Mg$ is completely consumed in the reaction, what mass of $MgO$ is produced? ($M_{Mg} = 24$, $M_O = 16$)
Ans
20 g
Review
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Calculate Moles of Mg
$n = m/M = 12/24 = 0.5$ mol
$$n_{Mg} = 0.5 \text{ mol}$$
Determine Moles of MgO
From the equation, 2 mol Mg → 2 mol MgO (1:1 ratio). So 0.5 mol Mg → 0.5 mol MgO.
$$n_{MgO} = 0.5 \text{ mol}$$
Convert to Mass
$m = n \times M = 0.5 \times 40 = 20$ g
$$m_{MgO} = 20 \text{ g}$$
8
$2Na + Cl_2 \rightarrow 2NaCl$. 23 g of sodium reacts with excess chlorine, yielding 50 g of $NaCl$. What is the percentage yield? ($M_{Na} = 23$, $M_{Cl} = 35.5$)
Ans
85.5%
Review
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Calculate Theoretical Yield
$n_{Na} = 23/23 = 1$ mol → 1 mol NaCl. $m = 1 \times (23 + 35.5) = 58.5$ g
$$\text{Theoretical yield} = 58.5 \text{ g}$$
Calculate Percentage Yield
$\%\text{ yield} = 50/58.5 \times 100$
$$\%\text{ yield} = 85.5\%$$
9
A reaction has a percentage yield of 60% and a theoretical yield of 50 g. What is the actual yield?
Ans
30 g
Review
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Rearrange the Formula
Actual yield = (Percentage yield / 100) × Theoretical yield
$$\text{Actual yield} = \frac{60}{100} \times 50$$
Calculate
$0.60 \times 50 = 30$ g
$$\text{Actual yield} = 30 \text{ g}$$
10
$Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$. 160 g of $Fe_2O_3$ reacts with 84 g of $CO$. What is the theoretical yield of iron? ($M_{Fe} = 56$, $M_O = 16$, $M_C = 12$)
Ans
112 g
Review
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Calculate Moles
$n_{Fe_2O_3} = 160/160 = 1$ mol, $n_{CO} = 84/28 = 3$ mol
$$n_{Fe_2O_3} = 1 \text{ mol}, \quad n_{CO} = 3 \text{ mol}$$
Verify Stoichiometric Ratio
The equation needs 3 mol CO per 1 mol $Fe_2O_3$ — exactly available. Both consumed completely.
$$\text{Stoichiometric ratio confirmed}$$
Calculate Theoretical Yield of Fe
1 mol $Fe_2O_3$ → 2 mol Fe. $m = 2 \times 56 = 112$ g
$$\text{Theoretical yield of Fe} = 112 \text{ g}$$
11
$CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$. If 50 g of $CaCO_3$ reacts with 73 g of $HCl$, what is the theoretical yield of $CaCl_2$? ($M_{CaCO_3} = 100$, $M_{HCl} = 36.5$, $M_{CaCl_2} = 111$)
Ans
55.5 g
Review
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Calculate Moles of Each Reactant
$n_{CaCO_3} = 50/100 = 0.5$ mol, $n_{HCl} = 73/36.5 = 2$ mol
$$n_{CaCO_3} = 0.5 \text{ mol}, \quad n_{HCl} = 2 \text{ mol}$$
Identify Limiting Reactant
0.5 mol $CaCO_3$ needs 1 mol $HCl$ (available: 2 mol). $CaCO_3$ is limiting.
$$\text{Limiting reactant: } CaCO_3$$
Calculate Theoretical Yield
0.5 mol $CaCO_3$ → 0.5 mol $CaCl_2$. $m = 0.5 \times 111 = 55.5$ g
$$\text{Theoretical yield of } CaCl_2 = 55.5 \text{ g}$$
12
$2Al + 3Cl_2 \rightarrow 2AlCl_3$. If 54 g of aluminium reacts with 213 g of chlorine gas, what mass of $AlCl_3$ is produced? ($M_{Al} = 27$, $M_{Cl} = 35.5$)
Ans
267 g
Review
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Calculate Moles of Each Reactant
$n_{Al} = 54/27 = 2$ mol, $n_{Cl_2} = 213/71 = 3$ mol
$$n_{Al} = 2 \text{ mol}, \quad n_{Cl_2} = 3 \text{ mol}$$
Verify Stoichiometric Ratio
The equation needs 3 mol $Cl_2$ per 2 mol $Al$ — exactly available. Both are consumed completely.
$$\text{Stoichiometric ratio confirmed}$$
Calculate Mass of Product
2 mol Al → 2 mol $AlCl_3$. $m = 2 \times (27 + 106.5) = 2 \times 133.5 = 267$ g
$$m_{AlCl_3} = 267 \text{ g}$$
13
$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$. When 16 g of $CH_4$ reacts with 48 g of $O_2$, 22 g of $CO_2$ is obtained. What is the percentage yield of $CO_2$? ($M_C = 12$, $M_H = 1$, $M_O = 16$)
Ans
66.7%
Review
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Calculate Moles
$n_{CH_4} = 16/16 = 1$ mol, $n_{O_2} = 48/32 = 1.5$ mol
$$n_{CH_4} = 1 \text{ mol}, \quad n_{O_2} = 1.5 \text{ mol}$$
Identify Limiting Reactant
1 mol $CH_4$ needs 2 mol $O_2$ but only 1.5 mol available. $O_2$ is limiting.
$$\text{Limiting reactant: } O_2$$
Calculate Theoretical Yield
1.5 mol $O_2$ produces $1.5/2 \times 1 = 0.75$ mol $CO_2$. $m = 0.75 \times 44 = 33$ g
$$\text{Theoretical yield} = 33 \text{ g}$$
Calculate Percentage Yield
$\%\text{ yield} = 22/33 \times 100 = 66.7\%$
$$\%\text{ yield} = 66.7\%$$
14
$Al + 3AgNO_3 \rightarrow Al(NO_3)_3 + 3Ag$. 27 g of $Al$ reacts with 680 g of $AgNO_3$. How many grams of $AgNO_3$ remain unreacted? ($M_{Al} = 27$, $M_{Ag} = 108$, $M_N = 14$, $M_O = 16$)
Ans
170 g
Review
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Calculate Moles
$n_{Al} = 27/27 = 1$ mol, $n_{AgNO_3} = 680/170 = 4$ mol
$$n_{Al} = 1 \text{ mol}, \quad n_{AgNO_3} = 4 \text{ mol}$$
Identify Limiting Reactant
1 mol $Al$ needs 3 mol $AgNO_3$ (available: 4 mol). $Al$ is limiting.
$$\text{Limiting reactant: } Al$$
Calculate Remaining $AgNO_3$
3 mol $AgNO_3$ consumed. Remaining = $4 - 3 = 1$ mol = $1 \times 170 = 170$ g
$$\text{Remaining } AgNO_3 = 170 \text{ g}$$
15
$4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O$. 34 g of $NH_3$ reacts with 64 g of $O_2$, producing 30 g of $NO$. What is the percentage yield of $NO$? ($M_N = 14$, $M_H = 1$, $M_O = 16$)
Ans
62.5%
Review
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Calculate Moles
$n_{NH_3} = 34/17 = 2$ mol, $n_{O_2} = 64/32 = 2$ mol
$$n_{NH_3} = 2 \text{ mol}, \quad n_{O_2} = 2 \text{ mol}$$
Identify Limiting Reactant
2 mol $NH_3$ needs $(5/4) \times 2 = 2.5$ mol $O_2$ (available: 2 mol). $O_2$ is limiting.
$$\text{Limiting reactant: } O_2$$
Calculate Theoretical Yield of NO
2 mol $O_2$ produces $(4/5) \times 2 = 1.6$ mol NO. $m = 1.6 \times 30 = 48$ g
$$\text{Theoretical yield} = 48 \text{ g}$$
Calculate Percentage Yield
$\%\text{ yield} = 30/48 \times 100 = 62.5\%$
$$\%\text{ yield} = 62.5\%$$
Question 1 of 15
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1
Which statement best describes the limiting reactant in a chemical reaction?