Answer:
68.1% is percent yield of the reaction
Explanation:
The reaction of methane with oxygen is:
CH₄ + 2O₂ → CO₂ + 2H₂O
Where 2 moles of oxygen react per mole of CH₄
Percent yield is:
Actual yield (28.2g CO₂) / Theoretical yield * 100
To solve this question we need to find theoretical yield finding limiting reactant :
Moles CH₄:
15.1g CH₄ * (1mol / 16.04g) = 0.9414 moles
Moles O₂:
81.2g * (1mol / 32g) = 2.54 moles
For a complete reaction of 0.9414 moles of CH₄ are needed:
0.9414 moles CH₄ * (2 mol O₂ / 1mol CH₄) = 1.88 moles of O₂. As there are 2.54 moles, O₂ is in excess and CH₄ is limiting reactant
In theoretical yield, the moles of methane added = Moles of CO₂ produced. That is 0.9414 moles CO₂. In grams = Theoretical yield:
0.9414 moles CO₂ * (44.01g / mol) = 41.43g CO₂
Percent yield: 28.2g CO₂ / 41.43g CO₂ * 100=
