In reactions involving more than one reactant, one has to identify first, of all the reactant, which is completely consumed (limiting reagent), one can identify the limiting reagent as follow

N_{2} + 3H_{2} → 2NH_{3}

Initial mole 5 12 —

If N_{2} is the limiting reactant moles of NH_{3} produced = 10.

If H_{2} is the limiting reactant moles of NH_{3 }produced

= (3/2) ×12 = 8.

The reactant producing the least number of mole of the product is the limiting reactant, hence H_{2} is the limiting reactant.

The limiting reactant can also be ascertained by knowing the initial number of equivalents or milli equivalents of each reactant.

The reactant with least number of equivalents or milli equivalent is the limiting reactant.

The equivalent methods to identify the limiting reactant used not require balancing of chemical equation.

__Gravimetric Analysis__

Calculation involving in mass – mass relationship

In general, the following steps are adopted in making necessary calculations:

1. Write down balanced molecular equation for the chemical change

2. Write down the no of moles below the formula of each of the reactant and product

3. Write down the relative masses of the reactants and the products with the help of formula below the respective formula. These shall be the theoretical amounts of reactants and product.

4. By the applications of unitary method, mole concept or proportionality method, the unknown factor or factors are determined.

**Calculation involving percent yield :**

In general, when a reaction is carried out in the laboratory we do not obtain actually the theoretical amount of the product. The amount of the product that is actually obtained is called the **actual yield.**

Knowing the actual yield and theoretical yield the percentage yield can be calculated as

**Weight Volume Relationship**

2Mg(s) + O_{2}(g) → 2MgO(s)

In the above reaction, one can find out the volume of O_{2} at STP required to react with 10 gm of Mg. The moles of Mg is 10/24 .

The moles of O_{2} required would be 1/2 the moles of Mg. Therefore moles of O_{2} = (1/2)x (10/24) .

Since 1 mole of a gas (ideal) occupies 22.4L at STP,

therefore (1/2)x (10/24) moles of O_{2} would occupy,

(1/2)x (10/24) x 22.4L

= 4.67L.

**Volume Volume Relationship**

Let us consider the reaction

H_{2}(g) + ½O_{2}(g) → H_{2}O(l).

We are given 10L of H_{2} at a given temperature and pressure. How many liters of O_{2} would react with hydrogen at the same temperature and pressure?

From the ideal gas equation [PV = nRT] it is clear that the volume of an ideal gas is directly proportional to its no. of moles. Therefore under the same conditions of P and T,

Since the molar ratio is 2:1 (H_{2}:O_{2}),

∴ the volume ratio would also be 2:1.

Therefore the volume of O_{2} required would be 5L.

On the other hand if we need to calculate the volume of O_{2} at a different T and P, then

and dividing we get