The combustion of methane can be described by the equation:
CH4 (g) + 2 O2 (g) -> CO2 (g) +2H20 (g)
If 22.0 mL of CO2 (g) were collected, what volume of H20 (g) also could be collected under the same conditions?
22.0 mL CO2 = how many moles. That is 22.0 mL x (1 mol/22,400 ml) = ? mol CO2. Now convert that to mole H2O.
?mole CO2 x (2 moles H2O/1 mole CO2) = ? mole CO2 x 2/1 and that x 22,400 mL/mol will convert to mL H2O.
There is a shortcut you can use when working will all gases. You can work in L and never convert to moles (notice with my work above that I divided by 22,400 mL/mol to convert moles CO2 to moles H2O, then multiplied by 22,400 to convert moles H2O to mL H2O.
The shortcut, using just mL is
22.0 mL CO2 x (2L H2O/1 L CO2) = 44.0 mL H2O(g)
To find the volume of H2O gas that could be collected under the same conditions, we need to use the stoichiometry of the reaction. The balanced equation shows that 1 mole of CH4 produces 2 moles of H2O.
Step 1: Calculate the moles of CO2 produced
To do this, we need to use the ideal gas equation:
PV = nRT
Where:
P = pressure (assumed constant)
V = volume of CO2 (22.0 mL, but convert it to liters, so 0.022 L)
n = moles of CO2 (what we want to find)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (assumed constant)
Since we have the volume and assume constant temperature and pressure, we can rearrange the equation to solve for n:
n = PV / RT
Step 2: Convert moles of CO2 to moles of H2O
From the balanced equation, we know that 1 mole of CH4 produces 2 moles of H2O. So in this reaction, the number of moles of H2O will be twice the number of moles of CO2.
n_H2O = 2 * n_CO2
Step 3: Convert moles of H2O to volume
Now we can use the ideal gas equation again to find the volume of H2O gas in liters:
V_H2O = n_H2O * RT / P
Step 4: Calculate the final answer
Plug in the values and solve for V_H2O:
V_H2O = (2 * n_CO2 * RT) / P
Remember that n_CO2 is the moles of CO2 that we calculated in step 1 and P is the assumed constant pressure.