Stoichiometry

• The word ‘stoichiometry’ is derived from two Greek words – stoicheion (meaning element) and metron (meaning measure).
• Stoichiometry deals with the calculation of masses (sometimes volumes also) of the reactants and the products involved in a chemical reaction.
• This is done using balance chemical equation.

Chemical Reaction

• A chemical reaction takes place whenever a chemical change occurs.
  • For example: – Magnesium (Mg) + Oxygen (O) à Magnesium oxide (MgO).
• Reactants are the substances which undergo chemical change in the reaction.
• Products are the new substances, formed during the reaction.
  • For example: – Magnesium (Mg) + Oxygen (O) à Magnesium oxide (MgO). In this equation Magnesium and Oxygen are reactants and Magnesium oxide is the product formed.

Balance Chemical Reaction

• According to law of conservation of mass; mass can neither be created nor be destroyed in a chemical reaction. That is, the total mass of the elements present in the products of a chemical reaction has to be equal to the total mass of the elements present in the reactants.
• Number of atoms on each element remains the same, before and after a chemical reaction.
  • For example:- (a) 2Mg + O₂ à 2MgO
  • (b) Zn + H₂SO₄ à  ZnSO₄ + H₂

Balance Chemical equation in Stoichiometry

• CH ₄(g) + 2O₂ (g) à CO₂ (g) + 2H₂O (g).
• The above reaction gives the information as follows:-
  • One mole of CH₄ (g) reacts with two moles of O₂ (g) to give one mole of CO₂ (g) and two moles of H₂O (g).
  • One molecule of CH₄ (g) reacts with 2 molecules of O₂ (g) to give one molecule of CO₂ (g) and 2 molecules of H₂O (g).
  • 16 g of CH₄ (g) reacts with 2×32 g of O₂ (g) to give 44 g of CO₂ (g) and 2×18 g of H₂O (g).

 Problem:-

Calculate the amount of water (g) produced by the combustion of 16 g of methane.

Answer:-

The balanced equation for combustion of methane is:

CH₄ (g) + 2O₂ (g) à CO₂ (g) +2H₂O (g)

(i) 16 g of CH₄ corresponds to one mole.

(ii) From the above equation, 1 mol of CH₄ (g) gives 2 mol of H₂O (g).

2 mol of water (H₂O) = 2 × (2+16) = 2 × 18 = 36 g

1 mol H₂O = 18 g H₂O ⇒ (18g H₂O)/ (1 mol H₂O) = 1

Hence 2 mol H₂O × (18g H₂O)/ (1 mol H₂O)

= 2 × 18 g H₂O = 36 g H₂O

Limiting Reagent

• In a chemical reaction, reactant which is present in the lesser amount gets consumed after sometime and after that no further reaction takes place whatever be the amount of the other reactant present. Hence, the reactant which gets consumed, limits the amount of product formed and is, therefore, called the limiting reagent.

Problem:-

50.0 kg of N₂ (g) and 10.0 kg of H₂ (g) are mixed to produce NH₃ (g).

Calculate the NH₃ (g) formed. Identify the limiting reagent in the production of NH3 in this situation.

Answer:-

A balanced equation for the above reaction is written as follows:

Calculation of moles:

N₂ (g) + 3H₂ (g) ⇌2NH₃ g

moles of N₂

= 50.0 kg N₂ × (1000 g N)/ (1 kgN₂) x (1molN₂)/ (28.0g N₂)

= 17.86×10² mol

moles of H₂

= 10.00 kg H₂ × (1000g H₂)/ (1kg H₂)/ (1molH₂)/ (2.016g H₂)

= 4.96×10³ mol

According to the above equation, 1 mol

N₂ (g) requires 3 mol H₂ (g), for the reaction.

Hence, for 17.86×10² mol of N₂, the moles of H₂ (g) required would be

17.86×10² mol N₂ × (3 molH₂ (g))/ (1mol N₂ (g)

= 5.36 ×10³ mol H₂

But we have only 4.96×10³ mol H₂. Hence, dihydrogen is the limiting reagent in this case. So NH₃ (g) would be formed only from that amount of available dihydrogen i.e., 4.96 × 10³ mol

Since 3 mol H₂ (g) gives 2 mol NH₃ (g)

(4.96×10³ mol H₂ (g)) × (2molNH₃ (g))/ (3molH₂ (g))

= 3.30×10³ mol NH₃ (g)

=3.30×10³ mol NH₃ (g) is obtained.

If they are to be converted to grams, it is done as follows:

1 mol NH₃ (g) = 17.0 g NH₃ (g)

3.30×10³ mol NH₃ (g) × (17.0 g NH₃ (g))/ (1mol NH₃ (g))

= 3.30×10³×17g NH₃ (g)

= 56.1×10³g NH₃

= 56.1 kg NH₃

Problem:-

Calculate the amount of carbon dioxide that could be produced when

(i) 1 mole of carbon is burnt in air.

(ii) 1 mole of carbon is burnt in 16 g of dioxygen.

(iii) 2 moles of carbon are burnt in 16 g of dioxygen.

Answer:

The balanced reaction of combustion of carbon can be written as:

(ii) According to the question, only 16 g of dioxygen is available. Hence, it will react with 0.5 mole of carbon to give 22 g of carbon dioxide. Hence, it is a limiting reactant.

(iii) According to the question, only 16 g of dioxygen is available. It is a limiting reactant.

Thus, 16 g of dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon dioxide.

Problem:-

In a reaction: – A + B₂ → AB₂

Identify the limiting reagent, if any, in the following reaction mixtures.

(i) 300 atoms of A + 200 molecules of B

(ii) 2 mol A + 3 mol B

(iii) 100 atoms of A + 100 molecules of B

(iv)  5 mol A + 2.5 mol B

(v) 2.5 mol A + 5 mol B

Answer:-

A limiting reagent determines the extent of a reaction. It is the reactant which is the first to get consumed during a reaction, thereby causing the reaction to stop and limiting the amount of products formed.

(i) According to the given reaction, 1 atom of A reacts with 1 molecule of B. Thus, 200 molecules of B will react with 200 atoms of A, thereby leaving 100 atoms of A unused. Hence, B is the limiting reagent.

(ii) According to the reaction, 1 mole of A reacts with 1 mole of B. Thus, 2 mole of A will react with only 2 mole of B. As a result, 1 mole of A will not be consumed. Hence, A is the limiting reagent.

(iii) According to the given reaction, 1 atom of A combines with 1 molecule of B. Thus, all 100 atoms of A will combine with all 100 molecules of B. Hence, the mixture is stoichiometric where no limiting reagent is present.

(iv) 1 mole of atom A combines with 1 mole of molecule B. Thus, 2.5 mole of B will combine with only 2.5 mole of A. As a result, 2.5 mole of A will be left as such. Hence, B is the limiting reagent.

 (v) According to the reaction, 1 mole of atom A combines with 1 mole of molecule B. Thus, 2.5 mole of A will combine with only 2.5 mole of B and the remaining 2.5 mole of B will be left as such. Hence, A is the limiting reagent.