a) nr de mmoli = 470.4 /22.4 =210 mmoli
niu CO2 / niu CH4 = 3/4
niu CO2 = 3 niuCH4 /4
3niuCH4/4 + niu Ch4 = 210
7 niuCH4/ 4= 210
niuCH4 =120 mmoli
=>niuCO2 = 90 mmoli
masa CH4 = 120 × 16 = 1920 mg
masa CO2 = 90 × 44 = 3960 mg
b) masa molara medie = 16 ×( 120/210 ) + 44 × (90/210)= 9.142 + 18.857 = 28
c) PV= niuRT
1 atm = 1.325 x 10^5 Pa
x atmn = 1.5195 x 10^5
P = 1.14 atm
1.14 x 820 = (120+90+9) × 0.082 × T
934.8 = 17.958 x T
T = 52.05 K
Ovservatie: am plecat de la premisa ca presiunea este in N/m2
d)
masa molara medie = 16 x (120/219) + 44 × (90/219) + y × (9/219)
28 = 8.76 + 18.08 + y × 0.04
y = 28
Gazul este N2