19,8 g oxid al C (carbon) ocupa un volum de 2214 ml la 27 grade si 5 atm. Care e masa moleculara si oxidul?

Răspuns :

pV = nRT <=> pV = (m/M)×R×T => M(oxid) = mRT/pV = (19,8×0,082×300)/(5×2,214) = 487,08/11,07 = 44g/mol

M(oxid) = 44g/mol => oxid = CO2