Answer :
Carbon dioxide formation involves 59.84 grams from this reaction.
- Write the balanced chemical equation for the combustion of methane:
CH₄ + 2O₂ → CO₂ + 2H₂O
- Calculate the moles of each reactant:
Molecular weight of methane (CH₄):
12 + 4(1) = 16 g/mol
Moles of CH₄ = 37.9 g / 16 g/mol
= 2.37 mol
Molecular weight of oxygen (O₂):
2(16) = 32 g/mol
Moles of O₂ = 87.0 g / 32 g/mol
= 2.72 mol
- Determine the limiting reactant:
The balanced equation shows 1 mole of CH₄ reacts with 2 moles of O₂.
Therefore, 2.37 moles of CH₄ would need 4.74 moles of O₂ to fully react.
Since we only have 2.72 moles of O₂, O₂ is the limiting reactant.
- Calculate the moles of CO2 produced:
According to the balanced equation, 2 moles of O₂ produce 1 mole of CO₂.
Thus, 2.72 moles of O₂ will produce:
Moles of CO₂ = 2.72 moles O2 × (1 mole CO₂/ 2 moles O₂)
= 1.36 moles CO₂
- Convert moles of CO₂ to grams:
Molecular weight of CO₂ = 12 + 2(16)
= 44 g/mol
Grams of CO₂ = 1.36 moles × 44 g/mol
= 59.84 grams
