Em tham khảo nha :

\(\begin{array}{l}
a)\\
Mn{O_2} + 4HCl \to MnC{l_2} + C{l_2} + 2{H_2}O\\
{n_{Mn{O_2}}} = \dfrac{{3,48}}{{87}} = 0,04mol\\
{n_{C{l_2}}} = {n_{Mn{O_2}}} = 0,04mol\\
{V_{C{l_2}}} = 0,04 \times 22,4 = 0,896l\\
H = \dfrac{{0,672}}{{0,896}} \times 100\%  = 75\% \\
b)\\
2Fe + 3C{l_2} \to 2FeC{l_3}(1)\\
Cu + C{l_2} \to CuC{l_2}\\
{n_{Cu}} = \dfrac{{0,64}}{{64}} = 0,01mol\\
{n_{CuC{l_2}}} = {n_{Cu}} = 0,01mol\\
{m_{CuC{l_2}}} = 0,01 \times 135 = 1,35g\\
{n_{C{l_2}(1)}} = 0,03 - 0,01 = 0,02mol\\
{n_{FeC{l_3}}} = \dfrac{2}{3}{n_{C{l_2}(1)}} = \dfrac{1}{{75}}mol\\
{m_{FeC{l_3}}} = \dfrac{1}{{75}} \times 162,5 = 2,17g\\
m = 1,35 + 2,17 = 3,52g
\end{array}\)