Question

calculate the volume of oxygen gad required for the complete combustion of 64g of CH4 at STP
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Answer 1

Question :

Calculate the volume of oxygen gas requires for the complete combustion of 64g of CH₄ at STP.

Solution :

Mass of methane = 64g

Molar mass of methane = 16g

♦ First of we have to write balanced chemical reaction of combustion of methane.

$\dag\:\underline{\boxed{\bf{\orange{CH_4+2O_2\:\longrightarrow\:CO_2+2H_2O}}}}$

Stoichiometry deals with the calculation of masses or moles of the reactants and the products involved in a chemical reaction.

» It is clear from the chemical reaction that 1 mole of CH₄ requires 2 moles of O₂.

Moles of CH₄ = 64/16 = 4

∴ 4 moles of CH₄ requires 4 × 2 = 8 moles of oxygen gas.

We know that 1 mole substance occupies 22.4 litre volume at STP.

➙ Volume = moles × 22.4

➙ Volume = 8 × 22.4

➙ Volume = 179.2 litre