p% = V.purx100/V.impur
=> V.p = V.imp.xp%/100 = 224x75/100
= 168 L acetilena pura
168 L n moli
C2H2 + 5/2O2 => 2CO2 + H2O
22,4 2
=> n = 168x2/22,4 = 15 moli CO2 in c.n.
dar in conditiile date cat va fi volumul..??
PV = nRT
=> V = nRT/P = 15x0,082x(273+27)/3
= 123 L CO2