Gaseous methane (CH4) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). If 28.2 g of carbon dioxide is produced from the reaction of 15.1 g of methane and 81.2 g of oxygen gas, Calculate the percent yield of carbon dioxide. Be sure your answer has the correct number of significant digits in it.

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=

they both have a y in them

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)