chem quiz please help
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The theoretical and percentage yield for the reaction are:
CH₄ + 2O₂ —> CO₂ + 2H₂O
Molar mass of CH₄ = 16 g/mol
Mass of CH₄ from the balanced equation = 1 × 16 = 16 g
Molar mass of O₂ = 32 g/mol
Mass of O₂ from the balanced equation = 2 × 32 = 64 g
Molar mass of CO₂ = 44 g/mol
Mass of CO₂ from the balanced equation = 1 × 44 = 44 g
SUMMARY
From the balanced equation above,
16 g of CH₄ reacted with 64 g of O₂ to produce 44 g of CO₂
From the balanced equation above,
16 g of CH₄ reacted with 64 g of O₂
Therefore,
20 g of CH₄ will react with = (20 × 64 ) / 16 = 80 g of O₂
From the above calculation, a higher mass (i.e 80 g) of O₂ than what was given (i.e 30 g) is needed to react completely with 20 g of CH₄.
Therefore, O₂ is the limiting reactant
From the balanced equation above,
64 g of O₂ reacted to produce 44 g of CO₂
Therefore,
30g of O₂ will react to produce = (30 × 44) / 64 = 21 g of CO₂
Percentage yield = (Actual / Theoretical) × 100
Percentage yield = (25 / 21) ×100
Percentage yield = 119%
Learn more about stoichiometry:
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