When 8.88 g of C6H10 were reacted with excess of O2, 18.9 g of CO2 were recovered. Given this information, what is the % yield of CO2?

Question

Macy Bradley

Chemistry

17 April, 02:24

When 8.88 g of C6H10 were reacted with excess of O2, 18.9 g of CO2 were recovered. Given this information, what is the % yield of CO2?

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Jace Guerra · Answer 1

66.2% yield

The balanced equation for the reaction is:

2C6H10 + 17O2 = = > 12CO2 + 10H2O

So for every 2 moles of C6H10 consumed, we should get 12 moles of CO2. Or to simplify, for each mole of C6H10, we should get 6 moles of CO2. Now let's calculate the molar mass of C6H10 and CO2 and then determine how many moles of each we really have.

Atomic weight carbon = 12.0107

Atomic weight hydrogen = 1.00794

Atomic weight oxygen = 15.999

Molar mass C6H10 = 6*12.0107 + 10*1.00794 = 82.1436 g/mol

Molar mass CO2 = 12.0107 + 2*15.999 = 44.0087 g/mol

Moles C6H10 = 8.88 g / 82.1436 g/mol = 0.10810337 mol

Moles CO2 = 18.9 g / 44.0087 g/mol = 0.429460538 mol

Since we had 0.10810337 moles of C6H10, we should have gotten

6*0.10810337 = 0.648620221 moles of CO2, but only got 0.429460538 moles. So let's divide the actual yield by the theoretical yield to get the percentage yield.

0.429460538 / 0.648620221 = 0.662114014 = 66.2%