a)

Đặt số mol của $\rm Fe$ là $\rm 2x\,\,\left( mol \right)$

Đặt số mol của $\rm Cu$ là $\rm y\,\,\left( mol \right)$

$\rm 2Fe\,\,+\,\,6{{H}_{2}}S{{O}_{4\,\,d,n}}\,\,\to \,\,F{{e}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}\,\,+\,\,3S{{O}_{2}}\,\,+\,\,6{{H}_{2}}O$

$\rm 2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,6x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3x\left( mol \right)$

$\rm Cu\,\,+\,\,2{{H}_{2}}S{{O}_{4\,\,d,n}}\,\,\to \,\,CuS{{O}_{4}}\,\,+\,\,S{{O}_{2}}\,\,+\,\,2{{H}_{2}}O$

$\rm y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y\,\,\left( mol \right)$

Hồn hợp $\rm Cu,Fe$ có $35,2\,\,\left( g \right)$

$\rm \Rightarrow 2x.56\,+\,64y=35,2$

$\rm \Rightarrow 112x+64y=35,2$

$\rm {{n}_{S{{O}_{2}}}}=\dfrac{17,92}{22,4}=0,8\,\,\left( mol \right)\,\,\,\Rightarrow \,\,3x+y=0,8$

Ta có hệ phương trình: $\rm \begin{cases}112x+64y=35,2\\3x+y=0,8\end{cases}\Leftrightarrow\begin{cases}x=0,2\,\,\,\left(mol\right)\\y=0,2\,\,\,\left(mol\right)\end{cases}$

Số $mol$ của $\rm Fe$ là $\rm 2x=2.0,2=0,4\,\,\left( mol \right)$

Số $mol$ của $\rm Cu$ là $\rm y=0,2\,\,\left( mol \right)$

b)

$\rm {{n}_{{{H}_{2}}S{{O}_{4}}}}=6x+2y=6.0,2\,\,+\,\,2.0.2=1,6\,\,\left( mol \right)$

$\rm {{C}_{M}}=\dfrac{n}{{{V}_{dd}}}\,\,\Rightarrow \,\,{{V}_{dd}}=n.{{C}_{M}}=1,6\,.\,7=11,2\,\,\left( l \right)$

$\rm D=\dfrac{{{m}_{dd}}}{{{V}_{dd}}}\,\,\Rightarrow \,\,{{m}_{dd}}=D\,.\,{{V}_{dd}}=1,28\,.\,11,2=14,336\,\,\left( g \right)$