Question

How many grams of carbon dioxide (CO2) are found

in a 11.9 L container under 1.5 atm and 20.0oC.

Answer 1

pV=nRT

p = 1,5atm = 1,5*101325 = 151987,5 Pa
V = 11,9l = 0,0119m3
n = ?
R = 8,314 J/mol*K
T = 20°C = 20+273 = 293K

pV=nRT
151987,5*0,0119 = n*8,314*293
1808,6513=n*2436,002 ||:2436,002
0,7425 = n >>> moles of CO2

1 mole of CO2 = 44g

1 mole of CO2--------------44g
0,7425 moles of CO2----------x
x = 32,67g of CO2