# Methane vs Methanol

Natural gas (mostly methane, ) is very abundant in many Middle Eastern oil fields. However, the costs of shipping the gas to markets in other parts of the world are high because it is necessary to liquefy the gas, which has a boiling point of 164 . One possible strategy is to oxidize the methane (in a 1:1 mole ratio) to methanol, , which has a boiling point of 65  and can therefore be shipped more readily.

What volume of methanol is formed if 3.14×1011  of methane at standard atmospheric pressure and 25  is oxidized to methanol? The density of  is 0.791 . Assume that the oxidation of methane to methanol occurs in a 1:1 stoichiometry.

### Re: Methane vs Methanol

Vital units missing from calculation which makes it unworkable

If 3.14×1011 is litres you can used PV=nRT to find n as long as 25 is a temperature.

ONce you know moles of methane, you know moles of methanol. As this is a liquid, find its mass using the mass of 1 mole of methanol (approx 12 +3 +16 +1 g)
Then use the density to find the volume, if you know the units of volume

### Re: Methane vs Methanol

Yeah, just noticed that the units didn't transfer- I will put it in again:

- 3.14 *10^11 L of methane
- 25 degrees Celsius
- density of CH3OH is 0.791 g/mL

### Re: Methane vs Methanol

Well follow the method I have given you

PV=nRT, so n =PV/RT so P=1, V= 3.14×1011, R = 0.082?057 L?atm? K?1?mol?1 and T = 298

n = no of moles of methane

This is equal to the number of moles of methanol = n

As this is a liquid, find its mass using the mass of 1 mole of methanol (approx 12 +3 +16 +1 g) = approx 32g
1 mole weighs approx 32g
n moles weighs  (32/1) x n grams = 32n grams methanol

0.791g of methanol occupies 1 mL
32n grams methanol occupies (1.0.791) x 32n  millilitres

Then use the density to find the volume, if you know the units of volume