Methane (CH4) and oxygen (O2) react to form carbon dioxide (CO2) and water (H2O) in a 2:1 ratio by volume. If 10.0 L of methane and 20.0 L of oxygen are mixed at the same temperature and pressure, which is the limiting reactant, and what volume of water vapor is produced?
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To determine the limiting reactant, we need to compare the number of moles of methane and oxygen.
From the balanced equation:
1 mole of CH4 produces 1 mole of CO2 and 2 moles of H2O.
Converting the volumes (10.0 L CH4 and 20.0 L O2) to moles using the ideal gas law, assuming temperature and pressure are constant, we get:
Moles of CH4 = 10.0 L * (1 mol/22.4 L) = 0.446 mol CH4
Moles of O2 = 20.0 L * (1 mol/22.4 L) = 0.893 mol O2
From the stoichiometry:
The ratio of moles of CH4 to O2 is 1:2.
Therefore, 0.446 mol CH4 corresponds to (0.446 mol CH4) * (2 mol H2O/1 mol CH4) = 0.892 mol H2O
Since the mole ratio of CH4 to H2O is 1:2, the limiting reactant is CH4, as it would produce less H2O than the available O2.
To find the volume of water vapor produced, we can use the ideal gas law:
Volume of H2O = 0.892 mol H2O * (22.4 L/1 mol) = 19.98 L
Therefore, the volume of water vapor produced is approximately 19.98 L.
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