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.

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?

Well, that sounds like quite the chemical transformation! Let's calculate the volume of methanol formed using a touch of mathematical tomfoolery.

First, we need to convert the amount of methane (in moles) to methanol (also in moles). Since the stoichiometry is 1:1, the number of moles of methane oxidized will be the same as the number of moles of methanol formed.

Given that the volume of methane is 2.86×10^11 molecules, we can find the number of moles of methane using Avogadro's number (6.022×10^23 molecules/mol). So, moles of methane = (2.86×10^11 molecules) / (6.022×10^23 molecules/mol).

Next, we convert the moles of methane to grams by multiplying it by the molar mass of methane (16.04 g/mol).

Now that we know the mass of methane, we can convert it to volume using the density of methane (0.791 g/mol). By dividing the mass of methane by the density, we get the volume of methane.

Since the stoichiometry is 1:1, the volume of methanol formed will be equal to the volume of methane.

So, if you gather up all the numbers and give them a good mathematical juggling, you should be able to calculate the volume of methanol formed!

To calculate the volume of methanol formed, we need to follow these steps:

Step 1: Determine the moles of methane (CH4) using the given quantity in scientific notation.
The quantity of methane given is 2.86×10^11 molecules.

To convert molecules to moles, we use Avogadro's number, which is 6.022×10^23 molecules/mol. So,

Number of moles of CH4 = (Number of molecules of CH4) / (Avogadro's number)

Number of moles of CH4 = (2.86×10^11) / (6.022×10^23)

Step 2: Determine the moles of methanol (CH3OH) formed assuming a 1:1 stoichiometry.
Since the oxidation of methane to methanol occurs in a 1:1 stoichiometry, the number of moles of CH4 will be equal to the number of moles of CH3OH formed.

Number of moles of CH3OH = Number of moles of CH4

Step 3: Calculate the mass of CH3OH formed.
The given density of CH3OH is 0.791 g/mol.

Mass of CH3OH = Number of moles of CH3OH × Molar mass of CH3OH

Step 4: Calculate the volume of CH3OH formed.
To calculate the volume, we need to use the density of CH3OH.

Density = Mass / Volume

Rearranging the equation:

Volume = Mass / Density

Now we can substitute the mass of CH3OH and density into the equation to find the volume.

I will perform the calculations for you.

oops sorri its

2.87*10^11L

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.

2.86 x 10^11 WHAT