initial : pV = nRT => n = 5×60/0,082×293 = 12,486 moli
final = pV = nRT => n = 5 moli
n consumat = 12,486-5 = 7,486 moli O2
b) C2H2 + 5/2O2 => 2CO2 + H2O
26g..........2,5 moli O2
xg..............7,486 moli
X = 77,8544g C2H2
c) p0V0/T0 = p1V1/T1 <=> 2×60/293 = 1×V1/273 => V1 = 273×120/293 = 111,8L
V1/V0 = 111,8/60 = 1,863 ori