n\(_{C_2H_4Br_2}\) = \(\frac{18,8}{188}\)= 0,1 (mol)
a,PTHH :
C\(_2\)H\(_4\) + Br\(_2\) \(\rightarrow\) C\(_2\)H\(_4\)Br\(_2\)
(mol) 0,1 ← 0,1
b, n\(_{hh}\) = \(\frac{8}{22,4}\) = \(\frac{5}{14}\) (mol)
\(\Rightarrow\) n\(_{CH_4}\) = n\(_{hh}\)- n\(_{C_2H_4}\) = \(\frac{5}{14}\) - 0,1 = \(\frac{9}{35}\) (mol)
m\(_{CH_4}\) = \(\frac{9}{35}\) * 16 \(\sim\) 4,1(gam)
m\(_{C_2H_4}\) = 0,1 * 28 = 2,8 (gam)
Vậy :
%m\(_{CH_4}\) = \(\frac{4,1}{4,1+2,8}\cdot100\%\) \(\sim\) 59,4 (%)
%m\(_{C_2H_4}\) = 100% - 59,4% \(\sim\) 40,6 (%)
