Question

Question #c07af

Answer 1

Assuming this is an ideal gas, recall,

`PV = nRT`

Hence,

`1"atm" * V = 3"mol" * (0.08206"L" * "atm")/("mol" * "K") * 273"K"`

`therefore V approx 70"L"`

is the volume of methane gas given your data.

Answer 2

67.2 litres

Explanation:

At S.T.P, any gas will occupy 22.4 liters[given that you have 1 mole of that gas]
Thus, 3 moles of methane will occupy `3 xx 22.4` liters or 67.2 liters

Answer 3

I get `68.1 \ "L"~~70 \ "L"`.

Explanation:

Since the change of `STP` values in `1982`, the new molar volume is found to be around `22.7 \ "L"`.

Sources:

https://en.wikipedia.org/wiki/Molar_volume

https://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure

If methane were to be an ideal gas, then `3` moles of methane at `STP` would occupy around

`22.7 \ "L"*3=68.1 \ "L"`

If we want to round it off to one significant figure, that will be

`68.1 \ "L" ~~ 70 \ "L"`