Question

Problema 11 va rog, de pe foaiae va rog mult

Answer 1

s = 10,8g

daer (gaz) = 2,483

a = 13,44 L CO2
b = 10,8g H2O 

d gaz = M/Mgaz => M=dgaz*Mgaz = 2,483*28,9 = 72g/mol

n = V/Vm = 13,44/22,4 = 0,6 moli CO2

m CO2 = 0,6*44 = 26,4g = a

%C = 300a/11s = 300*26,4/11*10,8 = 66,66
%H = 100b/9s = 100*10,8/9*10,8 = 11,11
%O = 100-(%C+%H) = 22,22

n C = %C*M/AC*100 = 66,66*72/1200 = 4 atomi C

n H = %H*M/AH*100 = 11,11*72/100 = 8 atomi H

n O = %O*M/AO*100 = 22,22*72/1600 = 1 atom O

F.M => C4H8O