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Stiochiometry,Stiochiometric Calculations and Limiting Reagents






StioChiometry

It deals with calculation of masses of the products and reactant in the chemical equal.

Chemical Equation

The method of representing a chemical reaction with the help of symbols and formula of the substances involved in it is known as a chemical equation.
The substances which combine or react are known as reactants.
The new substances produced in a reaction are known as products.
A chemical equation is a short-hand method of representing a chemical reaction.
A balanced chemical equation has an equal number of atoms of different elements in the reactants and products.
Example of Balanced Chemical equation
C3H8 + 5O2 -> 3CO2 + 4H2O
C3H8 and O2 are reactant and CO2 and H2 O are product.
The coefficient 1 in C3H8 and 5 in O2 are called stoichiometric coefficients

Limiting Reagents/Reactants


The reactants or reagent which totally get consumed during the reaction and limits the reaction is termed or Limiting Reagent.

Solved Examples

Question 1
Balanced Chemical  Reaction is given below
X  + 2Y = Z
(i) If 5 moles of both X & Y are taken initially give the limiting reagent.
 
(ii) 200 atoms of X and Y both are taken
Solution
i) X  + 2Y = Z
So 1 mole of X and    2 mole of Y gives  1 mole of Z
 1/2  mole  of X    react with 1 mole of  Y
Therefore, 5/2  mole of X   reacts with 5 mole of Y
So, “Y” is the limiting reagent.
  
ii) 1 atom of X  reacts with 2 atom of Y
1/2 atom  of X  reacts with   1 atom of Y
200×1/2  atom        200 atoms
100 atoms  200 atoms
‘X’ is left with 100 atoms.
Therefore ‘Y’ is the limiting agent.
 
Question 2
Balanced Chemical  Reaction is given below
A2 + B = 2P
If 200 moles of both A2 & B are taken initially give the limiting reagent.
Solution
1 mole of A2 reacts with 1 mole of B
Therefore 200 mole of A2 reacts with 200 mole of B   
No, limiting reagents.
Question 3
Give the amount of Ammonia formed on reaction of 10 gm of each N2 and H2. Give the limiting reagent.
N2   +     3H2 = 2NH3
Solution
From the reaction ,it is clear
28 gm of N2 reacts with 6 gm of H2
1 gm  of N2 reacts with 6/28 = 3/14 grams of H2
So, for 10 gm = 30/14=15/7=2.1 gm of H2
Therefore ,H2 is left i.e. 10 – 2.1 gm
   = 7.9 gm
From N2 ammonia formed will be:-
N2        = NH3
28 gm of N2 makes  34 g of Amonia
So 10 gm = 34/28  ×10 gm =12.14 gm

Question 4
A2 + B2 = C. If 10 moles of each is taken give limiting reagent.
Solution
No limiting agent
Question 5
2A + 4B = p. If 200 atoms of each are taken give limiting reagent.
Solution
A is the limiting reagent

Reactions in Solutions

Majority of the experiments in lab are carried out in the forms of solution. So, let’s study some important measure of the substance in solutions like Mole fraction ,Molarity on next page


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Class 11 Maths Class 11 Physics Class 11 Chemistry

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