\(\begin{array}{l}
a)\\
mhhA = 48,95 - 25,825 = 23,125g\\
nA = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
hh:C{l_2}(a\,mol),{O_2}(b\,mol)\\
a + b = 0,4\\
71a + 32b = 23,125\\
\Rightarrow a = 0,265;b = 0,135\\
\% VC{l_2} = 33,75\% \\
\% V{O_2} = 66,25\% \\
b)\\
hh:Fe(a\,mol),Zn(b\,mol)\\
65a + 56b = 25,825\\
3a + 2b = 1,07\\
a = 0,22;b = 0,21\\
\% mFe = 47,7\% \\
\% mZn = 52,3\%
\end{array}\)
