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Limiting reagent refers to the reactant which gets completely dissolved during a chemical reaction. The product quantity produced due to this kind of reaction is very low since the chemical reaction gets stunned due to less availability of the reagent.
- Limiting reagent is also called limiting agents or limiting reactants.
- When a reaction is carried out practically, it is difficult to determine the exact proportions of the reactant.
- This is due to the limitations found in the measuring instruments.
- If one or more reagents which are present in excess quantities react with the limiting reagent, they are called excess reagents.
- The result of the reaction can be optimized when determining the quantity of the limiting reactants.
Key Terms: Limiting Reagent, Chemical Reaction, Limiting Agents, Limiting Reactants, Excess Reagents, Reagents, Mole Ratio, Product Yield Method
What is a Limiting Reagent?
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The process in which two or more substances react with each other and convert into new substances is called chemical reaction. The reactants are the substances that react and products are the new substances that are produced after the reaction.
- When these reactions are carried out, the product yield will be determined by the reactant that is completely consumed.
- The reactant which gets used up first thus stopping the reaction from proceeding further, and limits the product formed is called limiting reagent.
- The factor determining the amount of reactant is the mole ratio and not the masses of reactants present.
- These limiting reagents in turn determine the time when the reaction stops.
- Since the product yield is determined from the reactants, limiting reagents should be identified in order to calculate the percentage and quantity of product yield.
Limiting Reagent ExamplesLet us consider a chemical reaction for the formation of Sulphur hexafluoride to understand the concept of limiting reagent: S + 3F2 = SF6 As per the Stoichiometry, 1 mole of Sulphur reacts with 3 moles of fluorine to form 1 mole of Sulphur Hexafluoride and therefore 3 moles of Sulphur reacts with 9 moles of fluorine to form 3 moles of Sulphur Hexafluoride. In this case, all the available Sulphur gets consumed and therefore it limits the further reaction. Hence Sulphur is the limiting reagent and fluorine is the excess reagent. The remaining three moles of fluorine are in excess and do not react. |
Limiting Reagent
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| Chapter Related Concepts | ||
|---|---|---|
| Ester Hydrolysis | Fehling Test | Iodoform Test |
| Hell-Volhard-Zelinsky Reaction | Friedel-Crafts acylation reaction | Butyric Acid |
How to find Limiting Reagent?
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Two methods are used to find the limiting reactant for a chemical reaction.
Reaction Stoichiometry Method
The reaction Stoichiometry Method involves the comparison of the mole ratios to calculate the value of the limiting reagent. It is also known as mole ratio approach. The steps involved in the reaction stoichiometry method are as follows:
- First, write the balanced chemical equation for the reaction.
- Then, determine the amount of moles of reactant of each type involved in the reaction.
- The mole ratio is calculated by dividing the amount of reactant by the respective stoichiometric coefficient in the balanced equation.
- Then, compare the mole ratios of each reactant to determine the smaller mole ratio.
Product Yield Method
The product yield method involves determining the limiting reactant by comparing the amount of product produced by each of the reactants. The steps involved in the reaction are as follows:
- Like the reaction stoichiometry method, first write the balanced chemical equation for the reaction.
- Next, convert the amount of each reactant produced in the reaction into the number of moles.
- Determine the moles of products produced from each reactant after assuming other reactants are present in excess.
- The reactant with the least amount of product will form the limiting reagent.
Limiting Reagent Examples
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The limiting reagent examples are as follows:
| Example: 50.0 kg of N2(g) and 10.0 kg of H2 (g) are mixed to produce NH3 (g). Calculate the NH3 (g) formed. In this process of production of NH3, find the limiting reagent. Solution: A balanced equation for the above reaction is written as follows: Calculation of moles: N2(g)+ 3H2(g) ⇔ 2NH3(g) moles of N2 = 50.0 kg N2 × 1000 g N2/ 1 kg N2 × 1molN2/ 28.0g N2 = 17.86×102 mol moles of H2 = 10.00 kg H2 × 1000 g H2/ 1 kg H2 × 1mol H2/ 2.016 g H2 = 4.96×103 mol According to the above equation, 1 mol N2 (g) requires 3 mol H2 (g), for the reaction. Hence, for 17.86×10^2 mol of N2, the moles of H2 (g) required would be = 17.86×102 mol N2 × 3 mol H2(g)/1mol N2(g) = 5.36 ×103 mol H2 But we have only 4.96×103 mol H2. Hence, dihydrogen is the limiting reagent in this case. So NH3(g) would be formed only from that amount of available dihydrogen i.e., 4.96 × 103 mol. Since 3 mol H2(g) gives 2 mole NH3(g) = 4.96×103 mol H2(g) × 2 mol NH2(g)/3mol H2(g) = 3.30×103 mol NH3(g) = 3.30×103 mol NH3(g) is obtained. To convert to gram below method is followed: 1 mol NH3(g) = 17.0 g NH3(g) 3.30×103 mol NH3(g) × 17.0 g NH3(g)/1mol NH3(g) = 3.30×103×17 g NH3(g) = 56.1×103 g NH3 = 56.1 kg NH3 Note: There are two ways to calculate the limiting reagent in a chemical reaction. One method is by comparing the mole ratio of the reactants present in a chemical reaction. Another method is by comparing the masses of the reactants present in a chemical reaction. The reactant which produces a smaller amount of product is the limiting reagent. |
Things to Remember
- Limiting reagents is a part of CBSE class 11 unit 1 Some Basic Concepts of Chemistry.
- This unit carries a total of 10 periods and 4 to 6 marks.
- The reagent is important in a reaction since it is completely consumed it determines when the reaction stops.
- Volume, Number of moles, mass, and the partial pressure of the reactant helps in determining the limiting reagent.
- Two limiting reagents are never possible in a single chemical reaction.
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Sample Questions
Ques. What is the limiting reagent? (3 marks)
Ans. In a chemical reaction limiting reagent is the reactant that is consumed first and prevents any further reaction from occurring. The amount of product formed during the reaction is determined by the limiting reagent. For example, let us consider the reaction of solution and chlorine.
2Na+Cl2→2NaCl
2Na atoms react with 1 Cl2 molecule. So, if we have 6Na moles, 3Cl2 molecules will be required, if there are more than 3 moles of Cl2 gas, sodium(Na) will act as a limiting reagent and some Cl2 molecules will remain as an excess reagent.
Ques. Why is limiting reagent important? (3 marks)
Ans. The limiting reagent is the reactant that is completely used up in a reaction and thus determines when the reaction stops. From the reaction stoichiometry, the exact amount of reactant needed to react with another element can be calculated.
- If the reactants are not mixed in the correct stoichiometric proportions (as indicated by the balanced chemical equation), then one of the reactants will be entirely consumed while another will be leftover.
- The limiting reagent is the one that is totally consumed; it limits the reaction from continuing because there is none left to react with the in-excess reactant. In chemistry, it is like the weakest link in a process, that limits how much reaction you can expect to occur.
Ques. Take the reaction: NH3 + O2 →NO + H2O. In an experiment, 3.25 g of NH3 are allowed to react with 3.50 g of O2.
a. Which reactant is the limiting reagent?
b. How many grams of NO are formed?
c. How much of the excess reactant remains after the reaction? (3 marks)
Ans. a. O2 reactant is the limiting reagent
- 2.63g of NO are formed
- 1.76g NH3 of the excess reactant remains after the reaction
Ques. If 4.95 g of ethylene (C2H4) is combusted with 3.25 g of oxygen.
a. What is the limiting reagent?
b. How many grams of CO2 are formed? (2 marks)
Ans. a. O2 is the limiting reagent.
- 2.98g CO2 are formed
Ques. Consider the reaction of C6H6 + Br2 →C6H5Br + HBr
a. What is the theoretical yield of C6H5Br if 42.1 g of C6H6 reacts with 73.0 g of Br2?
b. If the actual yield of C6H5Br is 63.6 g, what is the percent yield? (2 marks)
Ans. a. 71.6g C6H5Br is the theoretical yield of C6H5Br if 42.1 g of C6H6 reacts with 73.0 g of Br2
- 88.8% is the percent yield if the actual yield of C6H5Br
Ques. A reaction container holds 5.77 g of P4 and 5.77 g of O2. The following reaction occurs P4 + O2 → P4O6. If enough oxygen is available then the P4O6 reacts further: P4O6 + O2 →P4O10.
a. What is the limiting reagent for the formation of P4O10?
b. What mass of P4O10 is produced?
c. What mass of excess reactant is left in the reaction container? (3 marks)
Ans. a. O2 is the limiting reagent for the formation of P4O10
b. 5.78g P4O10 of mass of P4O10 is produced
c. 5.76g P4O6 remain is left in the reaction container
Ques. How do you calculate the mass of zink and iodine that are consumed to produce zink iodide? (2 marks)
Ans. In order to calculate the mass of zink and iodine, write the balanced reaction Zn + I2 → ZnI2. On the basis of this equation, 1 mole of Zink and 1 mole of Iodine react with each other in order to produce zink iodide. This forms the primary base for stoichiometric ratios.
Ques. What is a limiting reactant? (2 marks)
Ans. A limiting reactant refers to a substance that takes part in a chemical reaction and helps to determine the total number of products that can be produced from the reactants when there is an absence of the reactants in stoichiometric qualities.
Ques. What two differences between a reagent and a reactant? (2 marks)
Ans. The differences between a reagent and a reactant are as follows:
| Reagents | Reactants |
|---|---|
| A reagent is a catalyst | A reactant is a substrate |
| It binds a reaction | It is consumed in a reaction. |
Ques. When 4.00 mol H2 is mixed with 2.00 mol Cl2 how many moles of HCL can form? H2(g) + Cl2(g) ⇔ 2 HCL. Calculate the moles of each reactant and find the limiting reagent? (3 marks)
Ans. Limiting Reagent using Moles:
HCL from H2:
4.00 mol H2 × 2.00 mol HCL / 1.00 mol H2 = 8.00 mol HCL (not possible)
HCL from Cl2:
2.00 mol Cl2 × 2.00 mol HCL / 1.00 mol Cl2 = 4.00 mol HCL (smaller number)
The limiting reagent is Cl2 since it is used up first. Thus, Cl2 produces a smaller amount of Product.
Checking Calculations:
| Category | H2 | Cl2 | HCL |
|---|---|---|---|
| Initially | 4.00 mol | 2.00 mol | 0 mol |
| Reacted | -2.00 mol | -2.00 mol | +4.00 mol |
| Left after reaction | 2.00 mol excess | 0 mol limiting | 4.00 mol |
Ques. If 4.80 mol Ca is mixed with 2.00 mol N2 which is the limiting reagent in 3Ca(s) + N2(g) ⇔ Ca3N2(s)? (3 marks)
Ans. Limiting Reagent using Masses:
Moles of Ca3H2 from Ca:
4.80 mol Ca × 1.00 mol Ca3N2 / 3.00 mol Ca = 1.60 mol Ca3N2 (Ca used up)
Moles of Ca3N2 from N2:
2.00 mol N2 × 1.00 mol Ca3N2 / 1.00 mol N2 = 2.00 mol Ca3N2 (not possible)
When 1.60 mol Ca3N2 forms, all Ca gets used up. Thus, Ca is the limiting reagent.
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