Answer:
2 liters of methanol will produce 4 liters of steam and 2 liters of carbon dioxide.
Step-by-step explanation:
We need to start with a balanced equation. The unbalanced reaction is:
CH3OH + O2 = H2O + CO2 [A combustion reaction]
This is a bit tricky to balance. Start with the most complex molecule and assign it a coefficient of "1:"
1CH3OH + O2 = H2O + CO2
One methanol brings 1 carbon atom (C), so lets assign that atom to the CO2 product molecule on the right:
1CH3OH + O2 = H2O + 1CO2
The methanol brings 4 hydrogen atoms. The only place hydrogen atoms can go on the product side is water, H2O. Let's change the coefficient for water to "2," to account for the 4 H atoms coming in from the methanol.
1CH3OH + O2 = 2H2O + 1CO2
The C and H atoms are accounted for, so lets look at the oxygen atoms.
As written above, methanol contributes 1 O atom and O2 brings 2, for a total of 3 O atoms. The product side has 4 O atoms. 2 from the 2 water molecules and 2 from the one CO2 molecule. This is a problem since the number is odd on the reactant side and even on the product side. The product side must have an even number of O atoms, since the product side will always have an even number.
Lets double the number of methanol molecules. That will provide 2 atoms of O.
2CH3OH + O2 = 2H2O + 1CO2
This will double the number of carbons, hydrogens, as well as the oxygens.. Lets find homes for the carbons first. Double the number of CO2 molecules:
2CH3OH + O2 = 2H2O + 2CO2
Carbons are balanced, so look at hydrogens next. 2 Methanol molecules brings in 8 H and 2 O. Since H2O is the only possible destination for the H atoms, make it's coefficient 4:
2CH3OH + O2 = 4H2O + 2CO2
The above equation produces molecules with a total of 8 oxygens (4 each on the 4H2) and 2CO2).. Two come from the methanol, and 2 more from the 1O2, leaving 4 unaccounted for. We only have one O2 molecule, so let's make that 3:
2CH3OH + 3O2 = 4H2O + 2CO2
It looks balanced:
Reactants Products
C 2 2
H 8 8
O 8 8
All atoms are accounted for: A Balanced Equation
2CH3OH(l) + 3O29(g) = 4H2O(g) + 2CO2(g) is balanced
Note: Steam (gas) and not liquid is the likely phase of the water in this exothermic reaction.
This equation tells us that 2 moles of methanol will combine with 3 moles of oxygen to form 4 moles of water and 2 moles of CO2.
The problem asks for the volumes of water vapor and carbon dioxide will be produced from 2.0 liters of methanol. A very handy constant from the gas las is the observation that 1 mole of ANY gas at standard temperature and pressure (STP) will occupy 22.4 liters of volume. While we are not given conditions of temperature or pressure, they are held constant. So we will be able to use the relative volumes of the gases instead of moles.
Therefore, rereading the balanced equation, we can say that 2 liters of methanol will produce 4 liters of steam and 2 liters of carbon dioxide.