Question

Un volum de metan se impurifica cu acetilena. Trecand 0,336 l amestec printr o solutie de brom se consuma 0,32 g brom. Care este compozitia in procente de volume a amestecului?

Answer 1

26g............320g
HC≡CH + 2Br2 ---> CH(Br)2-CH(Br2)
x=0.026g.....0,32g

n=m/M=0.026/26=0.001 moli C2H2

n=V/Vm==>V=n*Vm=0.001*22,4=0.0224 L C2H2

VCH4=0,336-0.0224=0.3136 L CH4

%C2H2=0.0224/0,0336*100=6,66%

%CH4=0.3136/0,336*100=93,33% 

Answer 2

Teorie:Bromul reactioneaza cu acetilena Calcule: C2H2 + 2 Br2 = C2H2Br4 1 mol acetilena reactioneaza cu 2*160 g brom x moli acetilena reactioneaza cu 0,32 g brom x=0.001 moli acetilena V=n*22,4=0,001*22,4=0,0224 l acetilena V metan=V total-V acetilena=0.336-0.0224=0.3136 l ℅CH4=v metan/v amestec*100=0.3136/0.336*100=93.33 ℅C2H2=v acetilena/v amestec*100=0.0224/0.336*100=6.666℅ Raspuns: In amdstec se afla 93.33℅ metan si 6.666℅ acetilena