If 6.0 liters of methane react with 12.0 liters of oxygen according to the equation
CH4 + 2O2 → CO2 + 2H2O, and the temperature and the pressure are kept constant, which of the following is true?
a. All the methane and oxygen react.
b. Oxygen is in excess.
c. Methane is in excess.
d. The volume of the container increases.
e. The products have a lower average kinetic energy
Let's convert 6.0 L methane to mols. (That isn't required but it's a place to start.) At standard T and P, gases occupy 22.4L; therefore,
mols CH4 = 6.0/22.4 = about 0.27 mols. (I'm rounding here to make it simpler.)
mols O2 = 12.0/22.4 = 0.54 mols.
The coefficients in the balanced equation tells us that 1 mol CH4 will use 2 mols O2. We don't have a mole but we have about 0.27 mol CH4. Therefore, 0.27 mols CH4 will use 2*0.27 = 0.54 mol O2. That's exactly the amount of O2 we have which means neither reagent will be in excess; i.e., each reagent reacts completely.
So a is true and b and c obviously are ot true. c isn't true because there an equal number of moles on each side )3 on the left and 3 on the right). d isn't true because the problem states that P and T don't change and K.E. is related to T.
So much for the basics. When ALL GASES make up the problem we can use a shortcut. The shortcut is that when all gases are involved we can use volume (liters) as if L = mols. Technically then we don't need to convert everything to mols. That means one quick look at the equation tells us that 6.0 mols CH4 will combine exactly with 12.0 mols O2 to form CO2 and water. (We're assume this is at a temperature such that H2O is a gas and not a liquid). Thus, if CH4 and O2 reacta exactly it follows that a is true, b and c and not true, etc.
We can go further and say that the volume of CO2 formed will be 6.0 liters and the volume of H2O (as a gas) will be 12.0 L. (Note that makes 18 L gas on the left forming 18 L gas on the right so the statement above about c is false.
Does this help?
CH4 + 2O2→ Co2 + 2H2o CH4=23g
To determine which statement is true, we need to use the stoichiometry of the chemical equation and the principles of the ideal gas law.
First, let's examine the stoichiometry of the reaction. The balanced equation shows that 1 molecule of methane (CH4) reacts with 2 molecules of oxygen (O2) to produce 1 molecule of carbon dioxide (CO2) and 2 molecules of water (H2O).
From the given volumes, we can determine the number of moles of methane and oxygen. Using the ideal gas law equation (PV = nRT), where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature, we can solve for the number of moles:
For methane:
n(CH4) = (6.0 L) / Vm(CH4)
For oxygen:
n(O2) = (12.0 L) / Vm(O2)
Here, Vm refers to the molar volume at a given temperature and pressure.
Since the temperature and pressure are kept constant, the molar volumes for both methane and oxygen would be the same. Therefore, we can ignore Vm and consider the volumes directly.
Using the stoichiometry of the chemical equation, we find that 1 mole of methane reacts with 2 moles of oxygen. So, if the number of moles of oxygen is twice the number of moles of methane, then all the methane should react, and there would be excess oxygen.
From the given volumes, if the volume of oxygen is double the volume of methane, then the statement "Oxygen is in excess" is true. Therefore, the correct answer is b. Oxygen is in excess.
To summarize, by comparing the stoichiometry of the reaction and the volumes provided, we can determine the limiting reactant, which in this case is methane. As a result, oxygen is in excess, and the correct answer is b.
a is the only one that is true.
Do you know why?