What volume is occupied by 21.0 g of methane (CH4) at 27 degrees Celsius and 1 atm?

The volume occupied by 21.0 g of methane (CH4) at 27 degree celsius and 1 atm is 32.3L

PV = nRT

Where;

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The volume is occupies by 21,0g of methane is 32, 3 L.

Explanation:

We use the formula PV=nRT. We convert the unit temperature Celsiud into Kelvin: 0°C=273 K---> 27°C= 27+273= 300K. We calculate the weight of 1 mol of CH4:

Weight 1 mol CH4= Weight C+ (Weight H)x4= 12 g+ 1g x4= 16 g

16 g---1 mol CH4

21g---x= (21g x 1 mol CH4)/16g= 1, 31 mol CH4

PV=nRT    > V= (nRT)/P

V= (1, 31 mol x 0,082 l atm/K mol x 300 K)/ 1 atm

V= 32, 3875 L

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