**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__
C

_{3}H

_{8 }+ 5O

_{2} -> 3CO

_{2} + 4H

_{2}O

C

_{3}H

_{8 }and O

_{2 }are reactant and CO2 and H

_{2} O are product.

The coefficient 1 in C

_{3}H

_{8 }and 5 in O

_{2} 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

A_{2} + B = 2P

If 200 moles of both A_{2} & B are taken initially give the limiting reagent.

**Solution**

1 mole of A_{2} reacts with 1 mole of B

Therefore 200 mole of A_{2} 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.

N_{2} + 3H_{2} = 2NH_{3}

Solution

From the reaction ,it is clear

28 gm of N_{2} reacts with 6 gm of H_{2}

1 gm of N_{2} reacts with 6/28 = 3/14 grams of H_{2}

So, for 10 gm = 30/14=15/7=2.1 gm of H_{2}

Therefore ,H_{2} is left i.e. 10 – 2.1 gm

= 7.9 gm

From N_{2} ammonia formed will be:-

N_{2} = NH_{3}

28 gm of N_{2} makes 34 g of Amonia

So 10 gm = 34/28 ×10 gm =12.14 gm

**Question 4**

A_{2} + B_{2} = 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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**Also Read**

Class 11 Maths
Class 11 Physics
Class 11 Chemistry