Answer
73.56 g CO₂
Procedure
To solve this exercise first consider the following ethanol combustion equation
C₂H₅OH(l)+3O₂(g)⟶2CO₂(g)+3H₂O(l)
Assuming that the oxygen supply is unlimited as there is no data regarding oxygen, ethanol will be our limiting reactant.
Convert from grams to moles to be able to use the equation
[tex]50g\text{ }C₂H₅OH\frac{1\text{ }mole}{46.07g}=1.085\text{ }mole\text{ }C₂H₅OH[/tex]Using the proportions given by the combustion equation
[tex]1.085\text{ }mole\text{ }C₂H₅OH\frac{2\text{ }mole\text{ CO}_2}{1\text{ }mole\text{ C^^^^2082H^^^^2085OH}}=2.17\text{ }mole\text{ CO}_2[/tex]Converting back to grams of CO2
[tex]2.17\text{ }mole\text{ }CO₂\frac{44.01g}{1\text{ }mole}=95.5285g\text{ }CO₂[/tex]Adjusting for the yield
[tex]95.5285gCO_{2\text{ }}\frac{77}{100}=73.56g\text{ }CO_2[/tex]