Calculate the amount of carbon dioxide that could be produced when 1 mole of carbon is burnt in air.

Question

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 1

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

      C +  O2 ----------------> CO2
     1       1                          1 

(i) As per the balanced equation, 1 mole of carbon burns in 1 mole of dioxygen (air) to produce 1 mole of carbon dioxide.

(ii) In this part mass of two reactants are given so we need to find the limiting reagent first

Atomic mass of C = 12 amu

Molecular mass of O₂ = 2 × 16.00 = 32 amu

According to Stoichiometry of the reaction

12 g C will react with 32 g of O₂

But we have only 16 g O₂

Given mass of O₂ (16 g) < required mass of O₂ (32g)

Hence O₂ is limiting reagent and product is always calculated with the help of mass of limiting reagent

Molecular mass of CO₂ = 12 + 2× 16 = 44 amu

From the reaction

32 g of O₂ produce = 44 g CO₂

1 g of O₂ Produce = 44/32 g CO2

16 g O₂ will produce = 44 × 16 / 32 = 22 g CO₂

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

In this part mass of two reactants are given so we need to find the limiting reagent first

Atomic mass of C = 12 amu

Molecular mass of O₂ = 2 × 16.00 = 32 amu

2 mol of C = 2 x 12 = 24 g

According to Stoichiometry of the reaction

12 g C will react with = 32 g of O₂

1 g of C will react with = 32/ 12 g O₂

24 g of C will react with = 24 × 32 / 12 = 64 g O₂

But we have only 16 g O₂

Given mass of O₂ (16 g) < required mass of O₂ (64g)

Hence O₂ is limiting reagent and product is always calculated with the help of mass of limiting reagent

Molecular mass of CO₂ = 12 + 2× 16 = 44 amu

From the reaction

32 g of O₂ produce = 44 g CO₂

1 g of O₂ Produce = 44/32 g CO2

16 g O₂ will produce = 44 × 16 / 32 = 22 g CO₂