Question #11272
How many grams of methane gas (CH4) are in a 15.8 liter sample at 1.2 atmospheres and 27°C?
Expert's answer
According to Ideal gas law:
n*8,314*(27 + 273) =
1.2*101325*15.8*10^-3
n = 0.77 mol
m = 0.77 mol * 16g*mol-1 =
12.32 g
n*8,314*(27 + 273) =
1.2*101325*15.8*10^-3
n = 0.77 mol
m = 0.77 mol * 16g*mol-1 =
12.32 g
