More chem help!!?

Question

how many grams of O2 are consumed as 100.0 L of CH4 are burned? the volumes of CH4 and O2 are measured at 25 degrees C and .925 atm. The molar mass of O2 is 32.00 g/mol

The balanced equation is:

CH4 + 2 O2 ---> CO2 +2 H2O

Answer 1

First, calculate the number of moles of CH4 using the ideal gas equation (remember to make sure your R corresponds to the pressure units you're using, so it's 0.08206 in this case, for atmospheres, and make sure your temp. is in Kelvins):
PV = nRT
(0.925atm)(100L) = (0.08206atmL/molK)(298.15K)n
92.5 = 24.466n
n = 3.781 moles CH4

So you have 3.781 moles of CH4 combusted.
Now, using the balanced chemical equation, we know that for every mole of CH4 combusted 2 moles of O2 are required. Therefore, the number of moles of O2 required is:
3.781 x 2 = 7.562 moles O2

Now convert the number of moles to grams by multiplying by the molar mass:
(7.562 moles)(32 g/mol) = 241.984g

Therefore 241.984g of O2 is consumed.

Answer 2

First you need to find moles of CH4 using the ideal gas equation:

PV = nRT, solving for n, we get:

n = PV/RT

Plugging in the values (in correct units!):

n = 0.925atm x 100.0L/0.08206Latm/moleK x (273 + 25)K = 3.782 moles

Now, use the balanced equation to get moles of oxygen required:

3.782mole CH4 x (2moles O2/1moleCH4) = 7.564 moles oxygen.

Now get grams oxygen by multiplying moles by molar mass for O2:

7.564moles x 32.00g/mole = 242.0g O2

Answer 3

PV=nRT
0.925 atm x 100.0 L = n x 0.08206 x 298 deg K
n = 3.783 moles of CH4
3.783 moles of CH4 requires 2x3.7826 moles of O2 = 7.565 moles of O2
7.565 moles O2 x 32 g/mol = 242.1 g O2