How many grams of carbon dioxide (CO2) are found
in a 11.9 L container under 1.5 atm and 20.0oC.

pV=nRTp = 1,5atm = 1,5*101325 = 151987,5 PaV = 11,9l = 0,0119m3n = ?R = 8,314 J/mol*KT = 20°C = 20+273 = 293KpV=nRT151987,5*0,0119 = n*8,314*2931808,6513=n*2436,002  ||:2436,0020,7425 = n >>> moles of CO21 mole of CO2 = 44g1 mole of CO2--------------44g0,7425 moles of CO2----------xx = 32,67g of CO2