Limiting Reagent – Definition
Limiting Reagent (also known as Limiting reactant) is the first reactant to be completely consumed in a chemical reaction. Once this reactant is over, the whole process of product formation stops. As this reactant limits how much product can be formed, this is a limiting reagent.
Excess Reagent (also known as Excess reactant) is a reactant in a chemical reaction that is not used up completely at the end of the reaction.
Here H2 is an excess reagent, and N2 is a limiting reagent.
What is a mole?
A mole is the SI unit of measurement used to measure the amount of any substance. Mole is written as mol.
1 mol = 6.023*1023 particles (particles can be anything from an atom to anything large).
Lets us learn some molar calculations
Molar calculations include gram to mole and mole to gram conversions.
Moles to grams: given moles/stoichiometric coefficient*gram molecular weight=grams…..(i)
Grams to moles: given weight(g)/gram molecular weight(g)*moles
Find the ratio of given weight(g)/gram molecular weight(g)=a constant
any of the substance whose moles you want to calculate (a)* stoichiometric coefficient= no of moles….(ii)
E.g. 1: How many grams are there in 2 moles of ammonia?
Applying equation (i) 2 moles/1moles*17g (stoichiometric coefficient 1 as 1 NH3 is there and gram molecular weight of NH3 is 17 of N + 3*1 of H=17 g)
=34 g of NH3.
E.g. 2: Calculate the number of moles of substances that are reacting with 39.0 g of N2 and H2 to form NH3
- N2(g)+3H2(g)→2NH3 (Balanced equation)
- Now let us find as per equation(ii)
- a=given weight i.e. 39.0 g/gram molecular weight of N2 i.e 28g=1.39
- no of moles of N2=1.39*1=1.39 moles
- no of moles of H2=1.39*3=4.17 moles
- no of moles of NH3=1.39*2=2.78 moles
What is the percentage yield?
Percentage yield can be defined as the number of moles of any product formed for the reactant used.
Why should we identify limiting reagents in a reaction?
Because limiting reagent helps to calculate the percentage yield of a reaction as the amount of product obtained is dependent on the limiting reagent added.
How to identify limiting reagents in a reaction?
When a chemical reaction is given, first we need to balance that equation, and then we can identify the limiting agent and also find out the excess reagents by following methods:
Method A: By comparing the mole ratios of the number of reactants consumed.
- It can be applied mainly to only two reactant systems
- In a two reactant system, say A and B, choose one reactant A and use a balanced chemical equation to find out the amount of other reactant B required to react with the first one A
- Presence of an excess amount of second reactant B than required shows that second reactant B is excess and first reactant A is the limiting reagent. In contrast, if the amount of second reagent B is less than required, then second reactant B will be the limiting reagent
- Balance the given chemical reaction (as per the law of conservation quantity of each element remains the same throughout a reaction, so the amount of element on the right and left sides remains the same)
- Convert masses into moles and compare mole ratios with the chemical equation
- If the amount of reactant is less than required, it is the limiting reagent, and the one present in excess becomes the excess reagent
Method B: By comparing the product amounts that can be formed from each of the reactants.
- It can be used in systems containing any number of reactants
- Balance the given chemical equation
- Convert all masses into moles
- Using Stoichiometry, calculate the amount of product by using each reactant individually
- Denote the reactant with the least amount of product as a limiting reagent
Let us understand these methods with an example
Example : 76.4 g of C2H3Br3 reacts with 49.1 g of O2 in a chemical reaction. Find the limiting reagent in this reaction?
- Write the balanced equation : 4C2H3Br3+11O2→8CO2+6H2O+6Br2
- This means 11 moles of molecular oxygen is required to react with 4 moles of C2H3Br3
- Calculate the amount of oxygen required for the other quantities of C2H3Br3 as follows:
- Let us find out how many moles of C2H3Br3 are present in 76.4 g of C2H3Br3
- 1 mol = 266.72 g so 76.4 g=? mol
- 76.4g*1 mol/266.72 g=0.286 moles of C2H3Br3
- Now same way, find how many moles of O2 are present in 49.1 g of O2
- 49.1g*1mol/32g=1.53 moles of O2
- Now considering method A :
- According to this 11 moles of molecular oxygen are required to react with 4 moles of C2H3Br3 so if we assume all oxygen is used 1.53 moles of O2 reacts with 1.53*4/11 moles of C2H3Br3 i.e. 0.556 moles of C2H3Br3.
- Here we have only 0.286 moles of C2H3Br3.
- Compare the mole ratio of O2 and C2H3Br3 required by the balanced equation with the mole ratio present; it is clear that C2H3Br3 is the limiting reagent.
- Using Method B
- Amount of CO2 product formed by using C2H3Br3 as reactant =76.4 g C2H3Br3* 1 mol C2H3Br3/266.72g C2H3Br3*8 mol CO2/4mol C2H3Br3*44.01 g CO2/1mol CO2=25.2 g CO2
- Amount of CO2 product formed by using O2 as reactant = 49.1 g O2*1 mol O2/32 g O2*8 mol CO2/11 mol O2*44.01 g CO2/1 mol CO2=49.1 g CO2
- Reactant with the least amount of product is limiting reagent, so C2H3Br3 is the limiting reagent.
Problems and FAQ’s
- Calculate the mass of magnesium oxide if 2.40 g of Mg reacts with 10.0g of O2? Find the limiting reagent and also find the excess reagent left over?
- Write the balanced equation as follows:
- Interpreting mass we get:
2 Mg=24.31*2=48.61 g Mg
O2=16 g O2
2MgO=(24.31+16)*2=80.6 g MgO
- Let us now see how much product can be formed with Mg as reactant:48.61g gives 80.6g MgO, so 2.40 g gives___MgO?
- Now let us see how much product can be formed with O2 as reactant: 32 g gives 80.6 g MgO, so 10.0 g gives___ MgO?
10.0g *80.6g/32g=25.2 g
- So as per theory, the reactant that gives the least amount of product is called a limiting reagent, so here Mg is the limiting reagent.
- Now to find out excess reagent leftover:48.61g of Mg need 32 g of O2 to complete the reaction and give the product.
- 2.40 g of Mg need___ O2 to complete reaction? Solving this, we get 2.40*32/48.61=1.57 g of O2 is required to complete the reaction
- Reacted O2 10.0 g -1.57 g=8.43 g is the excess amount of O2 in the reaction.
- 56 g of N2 reacts with 30g of H2 to form NH3. Identify the limiting reagent and calculate the no of moles of NH3 formed
- .Writing balanced equation, we get N2+3H2→2NH3
- Stoichiometric coefficients: 1:3:2
- No of moles of N2 (nN2)=56/28=2
- No of moles of H2 (nH2)=30/2=15
- No of moles/ Stoichiometric coefficient
- Writing mole-mole relation of reactants:nN2/1:nH2/3 i.e 2 for N2 and 5 for H2 so N2 is the limiting reagent
- Now writing mole-mole relation of limiting reactant and product as below
- nN2/1=n NH3/2=2/1*2=n NH3
- No moles of NH3 formed=4 moles.
- Find the limiting reagent.
Substances can only react in a chemical reaction as long as the reactants are not used up.