Em tham khảo nha :

\(\begin{array}{l}
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} \to 6{H_2}O\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
{n_{{H_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,25mol\\
{m_{Fe}} = 0,25 \times 56 = 14g\\
{m_{Cu}} = 26,8 - 14 = 12,8g\\
{n_{Cu}} = \dfrac{{12,8}}{{64}} = 0,2mol\\
{n_{F{e_2}{{(S{O_4})}_3}}} = \dfrac{{{n_{Fe}}}}{2} = 0,125mol\\
{m_{F{e_2}{{(S{O_4})}_3}}} = 0,125 \times 400 = 50g\\
{n_{CuS{O_4}}} = {n_{Cu}} = 0,2mol\\
{m_{CuS{O_4}}} = 0,2 \times 160 = 32g\\
m = 50 + 32 = 82g\\
{n_{S{O_2}}} = \dfrac{3}{2}{n_{Fe}} + {n_{Cu}} = 0,575mol\\
{V_{S{O_2}}} = 0,575 \times 22,4 = 12,88l\\
b)\\
{n_{{H_2}S{O_4}}} = {n_{Fe}} = 0,25mol\\
{n_{{H_2}S{O_4}cd}} = \dfrac{{0,25 \times 120}}{{100}} = 0,3mol\\
{m_{{H_2}S{O_4}cd}} = 0,3 \times 98 = 29,4g\\
C{\% _{{H_2}S{O_4}}} = \dfrac{{29,4}}{{200}} \times 100\%  = 14,7\% 
\end{array}\)

Đáp án+Giải thích các bước giải:

$a,\\ n_{H_2}=\dfrac{5,6}{22,4}=0,25(mol)\\ Fe+H_2SO_4\ (l)\to FeSO_4+H_2\\ n_{Fe}=n_{H_2}=0,25(mol)\\ m_{Fe}=0,25.56=14(g)\\ m_{Cu}=26,8-14=12,8(g)\\ n_{Cu}=\dfrac{12,8}{64}=0,2(mol)\\ 2Fe+6H_2SO_4\to Fe_2(SO_4)_3+3SO_2+6H_2O\\ Cu+2H_2SO_4\to CuSO_4+SO_2+2H_2O\\ n_{Fe_2(SO_4)_3}=\dfrac{1}{2}.n_{Fe}=0,125(mol)\\ n_{CuSO_4}=n_{Cu}=0,2(mol)\\ m=0,125.400+0,2.160=82\\ n_{SO_2}=\dfrac{3}{2}.n_{Fe}+n_{Cu}=0,575(mol)\\ V=0,575.22,4=12,88(l)\\ b,\\ n_{H_2SO_4\ l}=n_{H_2}=0,25(mol)\\ n_{H_2SO_4\ bđ}=0,25.120\%=0,3(mol)\\ m_{H_2AO_4\ bđ}=0,3.98=29,4(g)\\ x=\dfrac{29,4}{200}.100\%=14,7$