Answer:
a) O2 is the limiting reactant
b) 75.70 grams CO2 (theoretical yield)
c) There remains 12.81 grams of C3H6O
d) The actual yield CO2 is 34.29 grams
Explanation:
Step 1: Data given
Mass of C3H6O = 46.3 grams
Mass of O2 = 73.2 grams
Molar mass of C3H6O = 58.08 g/mol
Molar mass of O2 = 32 g/mol
Step 2: The balanced equation
C3H6O + 4O2 → 3 CO2 + 3H2O
Step 3: Calculate moles C3H6O
Moles C3H6O = mass C3H6O / molar mass C3H6O
Moles C3H6O = 46.3 grams / 58.08 g/mol
Moles C3H6O = 0.793 moles
Step 4: Calculate moles O2
Moles O2 = 73.2 grams / 32 g/mol
Moles O2 = 2.29 moles
Step 5: Calculate limiting reactant
For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O
O2 is the limiting reactant. It will completely be consumed. (2.29 moles).
C3H6O is in excess. There will react 2.29/4 = 0.5725 moles C3H6O
There will remain 0.793 - 0.5725 = 0.2205 moles C3H6O
This is 0.2205 moles * 58.08 g/mol =<u> 12.81 grams</u>
Step 6: Calculate moles of CO2
For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O
For 2.29 moles O2 we need 3/4 * 2.29 = 1.72 moles CO2
This is 1,72 moles * 44.01 g/mol = <u>75.70 grams CO2</u>
Step 7: Calculate actual yield
% yield = 45.3 % = 0.453 = (actual yield / theoretical yield)
actual yield = 0.453 * 75.70 = <u>34.29 grams</u>
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