LIMITING REACTANT (LIMITING REAGENT)

The limiting reactant is the reactant in a chemical reaction that is completely consumed or used first. Once this reactant is consumed, the reaction stops, and no more products can be formed, even if other reactants are still available in excess. This reactant determines the maximum yield of a reaction.

EXCESS REACTANT (EXCESS REAGENT)

The excess reactant is a substance that is not completely consumed or used in a chemical reaction. Some of these reactants remain unreacted after the completion of the reaction. This reactant does not limit or determine the maximum yield of a reaction.

Consider the following reaction of 2H₂ + O₂ → 2H₂O.  In this reaction when 1 mole of O₂ and 1 mole of H₂ are mixed all the H₂ will react completely and O₂ will left unreacted because for 1 mole of H₂,  ½  mole of O₂ is required which is available but for a complete reaction of 1mole of O₂, 2 moles of H₂ are required which are non available that is why H₂ is limiting reactant and O₂ is non limiting or excess reactant. 

IDENTIFICATION OF LIMITING REAGENT

To identify the limiting reagent, the following steps are involved:

1. Calculate the number of moles of all the reactants by using the information given in the question
2. Calculate the number of moles of products by comparing them with the number of moles of reactants using the balanced equation.
3. Identify the reactant that produced the least number of moles of product. That will be the limiting reagent.

IMPORTANT NUMERICALS  

Q: Reaction: 2N₂ + 3 H₂ → 2 NH₃. With 5 mol N₂ and 12 mol H₂, what is the limiting reactant, and how many mol of NH₃ are formed?

Sol:  From the balanced equation, it is obtained that the mole ratios of N₂ and NH₃ are 1:1. So, 5 moles of N₂ = 5 moles of NH_3.

Similarly, on comparing the number of moles of H₂ and NH₃, it is obtained that 1 mole of H₂ = 2/3 mole of NH₃.  The 12 moles of H₂ = (2/3) x12 = 8 moles of NH₃.

As from N₂ a lesser number of moles of product (NH₃) is obtained, therefore N₂ is the limiting reagent.

 Q: 4.5 g of aluminium reacts with 4.0 g of chlorine gas. The actual yield of AlCl₃ is 10.0 g. Find the theoretical yield and percent yield. Reaction: 2Al+3Cl₂→2AlCl₃

Sol: In this question, the values in g of two reactants are given. So, first of all, we have to calculate the exact number of moles of reactants and then of products. Thus, moles of Al = 4.5 / 27 = 0.1667 mol and moles of Cl₂ = 4.0 / 71 = 0.0563 mol.

From the balanced equation, it is indicated that the mole ratios of 2Al and 2AlCl₃ are 1:1.Thus, the number of moles of AlCl₃ formed by Al is 0.1667. whereas the number of moles of AlCl₃ formed by Cl₂ is 0.037= (2/3) x 0.0563). Thus, the number of moles formed by Cl₂ is 0.037, which is a limiting reactant (produces a lesser number of moles of product). Thus, to calculate theoretical yield that number of moles will be used which are lesser.

Now, theoretical yield (g) = moles x molar mass ⇒ 0.037 × 133.5 = 5.01 g

Percent yield = actual yield/ theoretical yield x 100 ⇒  (10.0 / 5.01) × 100 = 199.6%

Q: A reaction involves the combustion of 100 g of propane (C₃H₈) with 250 g of oxygen (O₂). Calculate the limiting reagent and the amount of excess reagent left over. The balanced equation is: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.

Sol: In this problem, firstly, the number of moles of reactants is calculated: for propane, the no. of moles = mass/molar mass ⇒ 100/44 = 2.27, for the oxygen molecule, the number of moles = 250/32 =7.81.

Now, compare both these moles with the moles of the product, say CO₂. From the balanced equation, the mole ratios of propane and carbon dioxide are 1:3. Thus the calculated number of moles of CO₂ are 3×2.27 = 6.81 moles, and for O₂, from the balanced equation, the mole ratios of O₂ and carbon dioxide are 5:3. Thus the calculated number of moles of CO₂ are (3/5) x7.81= 4.68 moles. As a smaller number of moles is produced by O₂, O₂ is the limiting reactant, and propane is the excess reagent.  

The amount of excess reagent can be calculated by comparing the number of moles of both reagents. From the balanced equation, it is obtained that the mole ratios of O₂ and propane are 5:1. Thus, the calculated mole ratios are 7.81: (1/5)7.81 =1.56. It is obtained that from 4.68 moles of propane, only 1.56 moles react with O_2, and 3.12 moles are left over.

The amount of excess reagent can be calculated by the mass-moles relationship. No. of moles = mass/molar mass ⇒ 1.56 =mass/44 ⇒ mass = 44×1.56 = 68.64 g of propane is used. amount left over = 100-68.64=31.36g.

Q: A complex reaction involves the reaction of 500 g of reactant A (molar mass = 100 g/mol) with 300 g of reactant B (molar mass = 50 g/mol) and 400 g of reactant C (molar mass = 80 g/mol). The balanced equation is: 2A + 3B + 4C → D + E. Calculate the limiting reagent, theoretical yield of D (molar mass = 200 g/mol), and amount of excess reagents left over.

Sol: In this problem, firstly, the number of moles of reactants are calculated: for A the no. of moles = mass/molar mass ⇒ 500/100 = 5 moles, for B the number of moles = 300/50 = 6 moles, and for C the number of moles = 400/80 = 5 moles.

Now, compare all these moles of reactants with the moles of product, say D. From the balanced equation, the mole ratios of A and D  are 2:1, B and D are 3:1, and C and D are 4:1, respectively. Thus, the calculated number of moles of D  by the reactant A is (1/2) x 5 = 2.5 moles. And from B, D has (1/3) x6 =2 moles, from C, D has (1/4) x 5 = 1.25 moles. From these values of D, it is indicated that the C reactant is the limiting reactant. Because it produces a lesser number of moles of product (D).

Theoretical yield (g) of D = moles x molar mass ⇒ 1.25x 200=250g.

 Now, compare the number of moles of other reactants with the limiting reactant to evaluate the amount of excess reagents left over.

4C :2A =1C = 2/4 A ⇒ 5C = (2/4)5A= 2.5A. Excess A = 5-2.5= 2.5 moles ⇒ mass = 2.5 x100 = 250g

Similarly. 4C = 3B ⇒ 1C = 3/4B ⇒ 5C = (3/4)5B =3.75B. Excess B = 6-3.75= 2.25 moles ⇒ mass = 2.25 x50= 112.5g.

Q: A sample contains 10 g of impure calcium reacting with 10 g of water. The calcium sample is 80% pure. Find the limiting reagent and mass of hydrogen gas formed. Ca+2H₂O→Ca(OH)₂+H₂

Soln: Mass of impure calcium = 10 g

Purity = 80%

Mass of pure calcium = 10 g x 0.8 = 8 g (mass of pure sample is calculated by multiplying the impure sample by its purity in fractions).

Moles of calcium = mass/molar mass =8 /40 = 0.2 mol

Moles of H₂O = 10/18 = 0.55

Moles of H₂ produced = 0.2 mol (since 1 mole of Ca produces 1 mole of H₂)

Compared with water, 2 moles of H₂O produce 1 mole of H₂. ⇒ 1 mole of H₂O =1/2 H₂ ⇒ 0.55 moles of  H₂O = (1/2)X 0.55 =0.277. Thus, a smaller number of moles is produced by calcium, which acts as the limiting reactant.

Mass of H₂ Produced:

Moles of H₂ = 0.2 mol

Molar mass of H₂ = 2 g/mol

Mass of H₂ = 0.2 mol x 2 g/mol = 0.4 g

Q: 5 g of Zn reacts with 100 mL of 1 M HCl. Some HCl remains unreacted and decomposes upon heating, giving off chlorine. Zn+2HCl→ZnCl2+H₂, 4HCl→2Cl₂+2H₂

Find:  Limiting reagent

Volume of H₂ at STP

Mass of Cl₂ formed from excess HCl decomposed

Sol: Moles of  Zn = 5 / 65.38 = 0.0765 mol
Moles of HCl = 0.1 L × 1 M = 0.1 mol (in solution, the number of moles can be calculated by applying the formula: No. of moles = molarity × volume in L). From the balanced equation, it is obtained that 1 mole of Zn: 2 moles of HCl. So, 0.0765 moles of Znrequired 0.153= (2 × 0.0765) moles of HCl to react. Thus,

Required HCl = 0.0765 × 2 = 0.153 mol
Available HCl = 0.1 mol ⇒ thus, HCl is limiting, and Zn is in excess.

H₂ from HCl = 0.1 / 2 = 0.05 mol (from the balanced equation, the mole ratio between HCl and H₂ is 2:1).
Volume at STP = 0.05 × 22.4 = 1.12 L (no. of moles = volume/22.414L)

No Cl₂ formed (no excess HCl left).

Q: 100 mL of 0.5 M Na₂CO₃ reacts with 150 mL of 1 M HCl. Find the volume of CO₂ released at STP and the moles of HCl left. Reaction is: Na₂CO₃+2HCl→2NaCl+CO₂+H₂O.
Sol: Moles of Na₂CO₃ = 0.1 × 0.5 = 0.05 mol (in solution, the number of moles can be calculated by applying the formula: no. of moles = molarity x volume in L) =0.5 x 0.1 = 0.05 moles.
Moles of HCl = 0.15 × 1 = 0.15 mol. From the balanced equation, it is obtained that 1 mole of Na₂CO₃: 2 moles of HCl. So, 0.05 moles of Na₂CO₃ required 0.1= (2 × 0.05) moles of HCl. Thus,

Needed HCl = 2 × 0.05 = 0.1 mol
Available HCl = 0.15 mol ⇒ HCl left = 0.15 – 0.1 = 0.05 mol ⇒HCl is excess

The number of moles of CO₂ formed is obtained by comparing the number of moles of Na₂CO₃ in the balanced equation (because Na₂CO₃ has a lesser number of moles). So, CO₂ formed = 0.05 mol ⇒ Volume = 0.05 × 22.414 = 1.12 L (no. of moles = volume/22.414L)
Q: Calculate molecules of H₂O produced by 500 molecules each of H₂ and O₂. Which one is the limiting reactant? How much excess reactant is left unreacted? What would happen if molecules of O₂ were doubled?  2H₂ + O₂ → 2H₂O

Sol: In this problem, firstly, the number of moles of reactants are calculated: for H₂ the no. of moles = no. of particles/6.02×10²³ (here the number of moles are calculated by moles-Avogadro’s relation, because in question the information about number of particles is provided) ⇒ 500/6.02×10²³  = 83.05×10^-23 moles, for O₂ the number of moles are also = 83.05×10^-23 moles calculated by using moles-Avogadro’s relation.

Now, compare all these moles of reactants with the moles of product (H₂O). From the balanced equation, the mole ratios of H₂ and H₂O are 2:2 (or 1:1), and for O₂ and H₂O, the ratios are 1:2, respectively. Thus, the calculated number of moles of H₂O by the reactant H₂ is 83.05×10^-23 moles. And from O_2, the calculated number of moles is 166×10^-23 moles.

  From this number of moles of H₂O, it is indicated that the H₂ reactant is the limiting reactant. Because it produces a lesser number of moles of H₂O. molecules of water = moles x 6.02×10²³⇒ 83.05×10^-23 x 6.02×10²³⇒ 500 molecules.

 Now, compare the number of moles of other reactant (O₂) with the limiting reactant (H₂) to evaluate the molecules of excess reactant left over.

2H₂: O₂ = 2:1 ⇒ 1H₂ :1/2O₂ ⇒ 83.05×10^-23 H₂ : 1/2 83.05×10^-23 O₂ = 41.5×10^-23 O₂ moles(number of particles used = 250)⇒ Excess reactant of O₂= 83.05×10^-23– 41.5×10^-23 = 41.5×10^-23moles ⇒ excess number of particles = moles x 6.02×10²³⇒ 41.5×10^-23 x 6.02×10²³ = 250.

If the molecules of O₂ are doubled (i.e., 1000 molecules or moles=166.1×10^-23), no effect will be produced on the number of molecules of H₂O, because for H₂O formation only 41.5×10^-23 moles of O₂ are required. However, the number of molecules of excess reagent increased to 750 (moles=166.1×10^-23-41.5×10^-23 =124.6 x10^-23).  No. of particles = moles x 6.02×10²³ ⇒ 124.6 x10^-23 x 6.02×10²³ = 750.