Question

200 grame de carbid se hidrolizeaza formand 56 litri de acetilena in conditii normale. Ce puritate are carbidul din cele 200 de grame?

Answer 1

Răspuns:

M,CaC2= 64g/mol

niu=m/M=200g/64g/mol=3,125molCaC2 impur

V,acetilena= 56l

niu=V/Vm=56 l/22,4 l/mol= 2,5 mol acetilena

ecuatia chimica, cu substante pure

1mol...........................1mol

CaC2  +  H2O-----> C2H2+  CaO

x..................................2,5mol

x=niu carbid pur= 2,5mol

p= n,px100/n,i= 2,5x100/3,125=80%

Explicație: