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45 votes
The rubbing alcohol i just bought is marked 77% CH,OH. If burned 50.0 grams of this alcohol, how many grams of carbon dioxide would form, assuming the other 23.00% is not organic?

User Max Jacob
by
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1 Answer

11 votes
11 votes

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


50g\text{ }C₂H₅OH\frac{1\text{ }mole}{46.07g}=1.085\text{ }mole\text{ }C₂H₅OH

Using the proportions given by the combustion equation


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

Converting back to grams of CO2


2.17\text{ }mole\text{ }CO₂\frac{44.01g}{1\text{ }mole}=95.5285g\text{ }CO₂

Adjusting for the yield


95.5285gCO_{2\text{ }}(77)/(100)=73.56g\text{ }CO_2

User Don Chambers
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