posted by beeni on .
What volume of methanol is formed if 2.86×10^11 of methane at standard atmospheric pressure and 25 degree celcius is oxidized to methanol? The density of CH3OH is 0.791g/mol . Assume that the oxidation of methane to methanol occurs in a 1:1 stoichiometry.
2.86 x 10^11 WHAT
oops sorri its
Are you sure the density of CH3OH is 0.791 g/mol and not 0.791 g/mL?
Convert 2.86 x 10^11 L CH4 to moles. You can use the PV = nRT. Don't forget to use Kelvin for 25 C.
Since the reaction is 1:1, moles CH4 will be the same as moles CH3OH.
Convert moles CH3OH to grams CH3OH. moles = grams x molar mass.
The problem asks for volume so use the density to convert from grams to volume. This is where the disputed value of 0.791 comes in. The volume units will be determined by that unit.
Do I have to use PV=nRT or can I convert from 2.86* 10^11L to moles. I am not getting this and also in last step do I have to multiply or divide by 0.791 and then again divide by 1000 to get answer in L?
The problem states that the 2.86 x 10^11 liters is at standard pressure and 25 C. That is NOT STP so you can not simply divide 2.86 x 10^11 by 22.4 and PV = nRT is the easy way to do it. You COULD divide by 22.4 to convert to moles at STP, then correct for the T being 25 C (298 Kelvin) and not 273 K but that's more work then PV = nRT.
mass = volume x density for the last part of your post.
You have mass and density, you convert to volume. I don't know the units of the volume because of the strange 0.791 g/mol unit you posted. If that should be g/mL, then the units of the density calcn will be mL and you can divide by 1000 to convert to liters.