For the combustion of ethane, beginning with 2.6 mol of CH3CH3,
2CH3CH3(g) + 7O2(g) → 4CO2(g) + 6H2O(g)
a. how many moles of O2 are required to completely consume the CH3CH3?
9.1 moles
b. how many moles of CO2 are obtained when the CH3CH3 is completely combusted?
5.2 moles
c. how many moles of H2O are obtained when the CH3CH3 is completely combusted?
7.8 moles
I agree
To determine the number of moles of O2 required to completely consume the CH3CH3 in the combustion reaction, you can use the stoichiometric coefficients from the balanced equation.
In this case, the stoichiometric coefficient for O2 is 7. According to the balanced equation, 2 moles of CH3CH3 react with 7 moles of O2 to produce 4 moles of CO2 and 6 moles of H2O.
So, for every 2 moles of CH3CH3, you will need 7/2 moles of O2.
To find the number of moles of O2 required to completely consume 2.6 moles of CH3CH3, you can set up a proportion:
(2.6 moles CH3CH3) / (2 moles CH3CH3) = (x moles O2) / (7/2 moles O2)
Solving this proportion will give you the number of moles of O2 required to completely consume the CH3CH3. In this case, the answer is 9.1 moles of O2.
To determine the number of moles of CO2 and H2O produced when the CH3CH3 is completely combusted, you can once again use the stoichiometric coefficients.
According to the balanced equation, for every 2 moles of CH3CH3, you will produce 4 moles of CO2 and 6 moles of H2O.
Using this information, you can set up proportions similar to the one shown above and solve for the number of moles of CO2 and H2O.
For CO2:
(2.6 moles CH3CH3) / (2 moles CH3CH3) = (x moles CO2) / (4 moles CO2)
Solving this proportion will give you the number of moles of CO2 obtained when the CH3CH3 is completely combusted. In this case, the answer is 5.2 moles of CO2.
For H2O:
(2.6 moles CH3CH3) / (2 moles CH3CH3) = (x moles H2O) / (6 moles H2O)
Solving this proportion will give you the number of moles of H2O obtained when the CH3CH3 is completely combusted. In this case, the answer is 7.8 moles of H2O.