Question

How many liters of oxygen at STP are required for the combustion of 4g of methane gas? Also, calculate the volume of CO₂ gas produced at STP.

Answer 1

Hint: 1 mole of any gas at S.T.P. occupies 22.4 L of volume

Step 1: CH₄ (g) + 2O₂ (g) 🡪 CO₂ (g) + 2H₂O (l)

1 mole 2 moles 1 mole 2 moles

1 mole of any gas at S.T.P. occupies 22.4 L of volume

Molar mass of CH₄ = 12 + 1 ✕ 4 = 16 g mol^-1

O₂ = 16✕ 2 = 32 g mol^-1

Step 2: Given mass of CH₄ = 4 g

Number of moles = Given mass / molar mass

Number of moles of CH₄ = 4/16 = ¼ mole

Step 3:As per the stoichiometry of the reaction

1 mole of CH₄ requires = 2 moles of O₂

¼ mole of CH4 requires = 2 ✕ ¼ moles of O₂

= ½ mole of O₂

1 mole of any gas at S.T.P. occupies 22.4 L of volume

_{½ mole of} O₂ = ½ ✕ 22.4 L

= 11.2 L of O₂

As per stoichiometry of the reaction

1 mole of CH₄ produces = 1 mol of CO₂

1/4 mol of CH₄ produces = ¼ mol of CO₂

1 mole of any gas at S.T.P. occupies 22.4 L of volume

_{1/4 mole of} CO₂ = 1/4 ✕ 22.4 L

= ¼ ✕ 22.4 = 5.6 L of CO₂

Final Answer :11.2 L of oxygen are required at STP for the combustion of 4 g of methane gas.

5.6 L of CO₂ gas will be produced at STP.