a is the only one that is true.
Do you know why?
No, I don't understand the concept.
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