The combustion of propane (C3H8) in your BBQ occurs by the following unbalanced reaction:
C3H8(g) + O2(g) CO2(g) + H2O(g) If 34.0 mL of CO2(g) is collected, what volume of H2O(g) also could be collected under the
same conditions?
Balance the equation.
C3H8 + 5O2 ==> 3CO2 + 4H2O
When using GAS reactions one may use volume (mL if you wish) as if volume = mols.
Therefore, 34.0 mL CO2 x (4 mols H2O/3 mols CO2) = 34 x (4/3) = ? mL H2O
45.3
To determine the volume of H2O(g) that could be collected under the same conditions, we need to use stoichiometry.
1. Write and balance the equation for the combustion of propane:
C3H8(g) + O2(g) -> CO2(g) + H2O(g)
2. Determine the stoichiometric ratio between CO2 and H2O. By examining the balanced equation, we can see that the ratio of CO2 to H2O is 1:1. This means that for every mole of CO2 produced, one mole of H2O is also produced.
3. Convert the given volume of CO2 to moles. Since we are given 34.0 mL of CO2, we need to convert it to moles using the ideal gas law:
PV = nRT
Assuming constant temperature and pressure, we can rearrange the equation to solve for moles:
n = PV / RT
4. Calculate the moles of H2O that would be produced. Since the stoichiometric ratio between CO2 and H2O is 1:1, the moles of H2O produced will be the same as the moles of CO2.
5. Convert the moles of H2O to volume. Use the ideal gas law again to convert the moles of H2O to volume:
V = nRT / P
Substitute the calculated moles of H2O, temperature, pressure, and the ideal gas constant to obtain the volume of H2O.
By following these steps, you should be able to determine the volume of H2O(g) that could be collected under the same conditions.