Propane (c3h8) reacts with oxygen in the ir to produce carbon dioxide and water. in particular experiment, 38.0 grams of carbon dioxide are produced from the reaction of 22.05 grams of propane with excess oxygen. whats is the porcent yield

Percentage yield = 57.57%

Explanation:

Given data:

Actual yield of carbondioxide = 38.0 g

Mass of propane = 22.05 g

Percentage yield = ?

Solution:

Chemical equation:

C₃H₈ + 5O₂   →   3CO₂ + 4H₂O

Number of moles of propane:

Number of moles of propane = mass / molar mass

Number of moles of propane = 22.05 g/ 44 g/mol

Number of moles of propane = 0.5 mol

Now we compare the moles of carbon dioxide with moles of propane.

                           C₃H₈                 :               CO₂

                              1                     :                 3

                             0.5                  :               3×0.5 = 1.5 mol

Mass of carbondioxide:

Mass of carbondioxide = moles × molar mass

Mass of carbondioxide = 1.5 mol × 44 g/mol

Mass of carbondioxide = 66 g

Percentage yield:

Percentage yield = actual yield / theoretical yield × 100

Percentage yield =  38 g/ 66 g × 100

Percentage yield = 57.57%

i believe the answer is three water molecules