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Răspuns :

puritatea = V pur/ Vimpur*100 => Vpur=80*400/100=320m³ C2H2

mt=320*100/75=426,666m³ C2H2

n=V/Vm=426,666/22,4=19 kmoli C2H2

HC≡CH + H2O ⇒ CH2=CH(OH) ---tautomerizare---> CH3-CH=OH (etanal)

nC2H2=nC2H5O=19 kmoli

MC2H5O=45kg/kmol

m=19*45=855kg C2H5O