Some Basic Concepts of Chemistry

1.4 In order to answer the question, we need to know the balanced equation for the combustion of carbon in dioxygen/air, which can be written as:

C (s) + O₂ (g) à CO₂ (g)

(i) We can see from the above equation,

1 mole of carbon reacts with 1 mole of oxygen to produce 1 mol of carbon dioxide.

In air, combustion is complete.

Therefore, CO₂ produced from combustion of 1 mole of carbon= Molar mass of CO₂= 44 g

(ii) As only 16 g of dioxygen is available, it can combine only with 0.5 mole of carbon, i.e., dioxygen is the limiting reactant.

Hence, CO₂ produced = 22 g

Here, dioxygen acts as the limiting reagent.

(iii) Here again, dioxygen is the limiting reactant.

16 g of dioxygen can combine only with 0.5 mole of carbon (even though 2 moles of carbon are available). Therefore, the CO₂ produced again is equal to 22 g.

1.1. (i) Molecular mass of H₂O = (2x Atomic mass of Hydrogen)+ Atomic mass of Oxygen

Atomic mass of Hydrogen = 1.008 amu

Atomic mass of Oxygen = 16.00 amu

So,

Molecular mass of H₂O = 2 (1.008 amu) + 16.00 amu =18.016 amu

(ii) Molecular mass of CO₂ = Atomic mass of Carbon + (2x Atomic mass of Oxygen)

Atomic mass of Carbon = 12 amu

Atomic mass of Oxygen = 16 amu

So,

Molecular mass of CO₂ = 12.01 amu + 2 x 16.00 amu = 44.01 amu

(iii) Molecular mass of CH₄ = Atomic mass of Carbon+ (4x Atomic mass of Hydrogen)

Atomic mass of Carbon = 12 amu

Atomic mass of Hydrogen = 1.008 amu

So,

Molecular mass of CH₄ = 12.01 amu + 4 (1.008 amu) = 16.042 amu