For the balanced equation shown below, if the reaction of 91.3 grams of C3H6 produces a 81.3% yield, how many grams of CO2 would be produced?

2C3H6+ 9O2 --> 6CO2+ 6H2O
233.3g

74.2g

81.3g

91.3g

Option A (233.3 gram)

Explanation:

Step 1: Data given

Mass of C3H6 produced = 91.3 %

% yield = 81.3 %

Step 2: The balanced equation

2C3H6+ 9O2 → 6CO2+ 6H2O

Step 3: Calculate moles C3H6

Moles C3H6 = mass C3H6 / molar mass C3H6

Moles C3H6 = 91.3 grams / 42.08 g/mol

Moles C3H6 = 2.17 moles

Step 4: Calculate moles CO2

For 2 moles C3H6 we need 9 moles O2 to produce 6 moles CO2 and 6 moles H2O

For 2.17 moles C3H6 we'll have 3*2.17 = 6.51 moles CO2

Step 5: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 6.51 moles * 44.01 g/mol

Mass CO2 = 286.5 grams

Step 6: Calculate actual yield of moles CO2

% yield = (actual yield / theoretical yield ) *100 %

actual yield = % yield * theoretical yield / 100 %

actual yield = (81.3 % * 6.51 moles ) / 100 %

actual yield = 5.29 moles

Step 7: Calculate actual yield of mass CO2

Mass CO2 = moles * molar mass

Mass CO2 = 5.29 * 44.01 g/mol

Mass CO2 = 233 grams

Option A (233.3 gram) is the clostest

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