Question

Ethanol, C,H,O, is added to gasoline to produce "gasohol,"a fuel for automobile engines. How many grams of O₂ arerequired for complete combustion of 421 g of ethanol?2CO₂(g) + 3H₂O(g)C₂H₂OH()+ 30₂(g)Ethanol

Answer 1

We have the ethanol combustion reaction to produce CO2 and water. The balanced equation of the reaction is:

$4C_2H_2OH+9O_2\rightarrow8CO_2+6H_2O$

We must first find the moles of ethanol corresponding to 421 grams. To do this we divide the mass by the molar mass of ethanol. The molar mass of ethanol is: 46.07g/mol. The moles of ethanol will be:

$molC_2H_2O=421gC_2H_2O\times\frac{1molC_2H_2O}{46.07gC_2H_2O}=9.14molC_2H_2O$

Now, by the stoichiometry of the reaction, we see that the Oxygen to Ethanol ratio is 9/4. So, the moles of oxygen needed will be:

$molO_2=9.14molC_2H_2O\times\frac{9molO_2}{4molC_2H_2O}=20.6molO_2$

We find the grams of oxygen by multiplying the moles by the molar mass. The molar mass of O2 is 31.9988g/mol.

$gO_2=20.6molO_2\times\frac{31.9988gO_2}{1molO_2}=658gO_2$

Answer: To complete the combustion of 421 g of ethanol are needed 658 grams of oxygen