If 20.0 L of methane, CH4, reacts with 200.0 L of oxygen, calculate the L of CO2 produced.?

Question

the given information is

CH4 (g) + 2 O2 (g) -> CO2 (g) + 2H2O (l)

Answer 1

The volume of any gas at any temp and pressure will contain the same number of moles as any other gas at the same temp and pressure.

eg: at standard temp and pressure (STP) the volume of 1 mole of any gas, is 22.4 L.

That means that in your reaction mixture you have ten times the number of moles of O2 as you do CH4. (20.0 L of CH4, 200.0 L of O2)

The balanced equation tells you that 1 mole of CH4 reacts with 2 moles of O2. This means that your limiting reagent is the CH4, because there is more then enough O2 to react with all the CH4.

The equation also tells you 1 mole of CH4 -----> 1 mole of CO2.

Since 1 mole of all gases has the same volume at the same temp and pressure then this means you will have 20.0 L of CO2.

Of course, this is as long as temp and pressure are held constant, I am assuming it is, since you have not provided any temp and pressure data