Every chemical reaction follows the laws of chemical combination. If a compound has a formula ABC, it is essential to have all the three components namely A, B and C to be present in the requisite proportions. It is obvious that the compound cannot be formed if either of the components is missing. In this topic we are going to discuss what will happen if all the three are present but not in requisite proportions.
When chemical reactions are carried out in the laboratory, we do not add reactants in the exact molar ratios indicated by the balanced equation. For various reasons one or more reactants is usually present in excess. In most cases, only one reactant is completely consumed at the end of the reaction, and this reactant determines the amount of products that can be formed. This is called limiting reagent or reactant. Once the limiting reagent is entirely consumed the reaction stops, the other excess reactants will still be present in some amount.
In a reaction where there is more than one reactant involved, the reactant that gets completely consumed and converted into the product is the limiting reactant.
For example in the case Magnesium ribbon burning in air to form MgO, Magnesium is the limiting reactant. In terms of mole concept the reactant which produces least number of moles of the product is the limiting reactant.
For example: In a reaction where 1mole of A and 2 moles of B combine to produce 4 moles of C, if one starts with 4 moles of A and 10 moles of B, the limiting reactant is A . This is explained in this way. since 1 mole of A produces 4 moles of C, 4 moles of A will produce 16 moles of C. But as per the reaction since 2 moles of B is what is required to produce 4 moles of C, 16 moles of C would require only 8 moles of B and not the entire 10 moles taken. Thus 2 moles of B will remain after complete exhaustion of A. Since A is the reactant that is completely converted to products, and is the controlling factor, A is the limiting reactant.
In terms of equivalences also one can define the Limiting reactant. "The reactant with the least number of equivalents or Milli-equivalents is the limiting reactant"
. The advantage of defining the limiting reactant in equivalences is that the equation need not be balanced to ascertain the limiting reactant.
Depending on the nature of the reaction the limiting reactant can be found. Finding the limiting reactant can be by calculating the number of moles present or by the strength in terms of Milli-equivalents or equivalents. If the reaction is an acid base reaction finding the limiting reactant can be done by suitable indicators. Generally it is done by volumetric method of analysis.
If the reaction is a precipitation reaction the stoppage of precipitation or incomplete precipitation gives the idea of the limiting reagent. If the reaction is dissolution, the clear solution, saturated or unsaturated will give the limiting reactant. In case of gaseous reactions the limiting reactants can be found by analyzing the mixture of gases that are formed.For chemical reactions involving two or more reactants, it is necessary to determine which one is the limiting reagent. The following procedure is helpful in determining the limiting reagent.
- Make sure that chemical equation is balanced.
- Calculate the number of moles of each of the reactants present in their given amount.
- Calculate the amount of products that can be formed from the complete reaction of each reactant.
- Determine which of the reactants would produce the least amount of the product. This is the limiting reagent.
Below you could see problems
An aluminium wire weighing 54g was dipped in 2L of 1M sulfuric acid. Find out which one of the reactants is the limiting reactant? ( Atomic mass of Al = 27) Solution:
The equation will be Al + H2SO4 → Al2(SO4)3 + H2 ↑
Balancing the equation we get 2Al + 3H2SO4 → Al2(SO4)3 + 3H2 ↑
The balanced equation suggests that 2 moles of Al requires 3 moles of H2SO4
54 g of Al is 54/27 = 2 moles which would require 3 moles of H2SO4.
Now we have 2L of 1M H2SO4 which is equal to only 2 moles.
∴ Sulfuric acid is the limiting reactant .
2.3 g of Sodium metal is lowered in to a 3L flask filled with chlorine gas at STP. Find the limiting reagent and the amount in grams or liters of the excess reagent present. At mass of Na = 23 , Cl = 35.5.
Balanced equation is 2Na + Cl2 → 2NaCl
2.3 g of sodium metal is 2.3/23 = 0.1 mole which require 1 X 0.1/2 = 0.05 moles of chlorine.
At STP 1 mole of chlorine gas occupies 22.4 L .
∴ 3 L contains 3/22.4 = 0.1339 moles.
Thus the limiting reactant is Sodium metal and excess reagent is Chlorine.
The amount of Cl2 present in excess is 0.1339 - 0.05 moles = 0.0839 moles which will occupy at STP
0 0839 X 22.4 / 1 = 1.88 L approx.
1. How many moles of which reactant will remain if 1.39 moles of N2 and 3.44 moles of H2 will react to form ammonia? find out how many grams of ammonia can be formed and how many moles of limiting reactant is required to completely exhaust the other reactant that is in excess?
2. If we take 1 mol of oxygen and 1 g of hydrogen in a reaction to make water, which will be the limiting reagent and how much of excess reagent is left behind?