Đáp án:

$a,\\ m=30,4\\ m_X=88(g)\\ b,\ 16(g)$

Giải thích các bước giải:

$a,\\n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)\\ n_{SO_2}=\dfrac{13,44}{22,4}=0,6(mol)\\ TN2:\\ Fe+2HCl\to FeCl_2+H_2\\ \xrightarrow{\text{Từ PTHH}}\ n_{Fe}=n_{H_2}=0,2(mol)\\ TN1:\\ 2Fe+6H_2SO_4\to Fe_2(SO_4)_3+3SO_2+6H_2O\ (*)\\ Cu+2H_2SO_4\to CuSO_4+SO_2+H_2O\ (**)\\ \xrightarrow{\text{Từ (*)}}\ n_{SO_2\ (*)}=\dfrac{3}{2}.n_{Fe}=0,3(mol)\\ \Rightarrow n_{SO_2\ (**)}=0,6-0,3=0,3(mol)\\ \xrightarrow{\text{Từ PTHH}}\ n_{Cu}=n_{SO_2\ (**)}=0,3(mol)\\ m_{Fe}=0,2.56=11,2(g)\\ m_{Cu}=0,3.64=19,2(g)\\ m=11,2+19,2=30,4\\ \xrightarrow{\text{Từ PTHH}}\ n_{Fe_2(SO_4)_3}=\dfrac{1}{2}.n{Fe}=0,1(mol)\\ \xrightarrow{\text{Từ PTHH}}\ n_{CuSO_4}=n_{Cu}=0,3(mol)\\ m_X=0,1.400+0,3.160=88(g)$

$b,\\ \xrightarrow{\text{Từ PTHH}}\ n_{FeCl_2}=n_{H_2}=0,2(mol)\\ NaOH+HCl\to Nacl+H_2O\\ FeCl_2+2NaOH\to Fe(OH)_2+2NaCl\\ 4Fe(OH)_2+O_2\xrightarrow{t^0} 2Fe_2O_3+4H_2O\\ \xrightarrow{\text{Từ PTHH}}\ n_{Fe_2O_3}=\dfrac{1}{2}.n_{FeCl_2}=0,1(mol)\\ m_{Fe_2O_3}=0,1.160=16(g)$

`n_{SO_2}=(13,44)/(22,4)=0,6(mol)`

`n_{H_2}=(4,48)/(22,4)=0,2(mol)`

Gọi x,y lần lượt là số mol `Fe,Cu`

Ta có quá trình oxi hóa khử :(khi hh X tác dụng dd `H_2SO_4` đặc)

$\text{$\mathop{Fe}\limits^{0}$→$\mathop{Fe}\limits^{+3}$+3e}$

$\text{x    → 3x}$

$\text{$\mathop{Cu}\limits^{0}$→$\mathop{Cu}\limits^{+2}$+2e}$

$\text{y      →2y}$

$\text{_________________________________}$

$\text{$\mathop{S}\limits^{+6}$+2e→$\mathop{S}\limits^{+4}$}$

$\text{.     1,2← 0,6(mol)}$

$\text{Bảo toàn e : 3x+2y=1,2(1)}$

Ta có PTHH:

`Fe+2HCl→FeCl_2+H_2`

⇒`x=n_{Fe}=n_{H_2}=0,2(mol)`

Thay vào (1) ta được : `y=0,3(mol)`

`⇒m=0,2.56+0,3.64=30,4(g)`

Ta có bảo toàn nguyên tố Fe:`n_{Fe_2(SO_4)_3}=1/2 n_{Fe}=1/2 .0,2=0,1(mol) ⇒m_{Fe_2(SO_4)_3}=0,1.400=40(g)`

`n_{CuSO_4}=n_{Cu}=0,3(mol)⇒m_{CuSO_4}=0,3.160=48(g)`

`⇒m_{m uối}= 40+48=88(g)`

b,PTHH:

`Fe_2(SO_4)_3+6NaOH→2Fe(OH)_3+3Na_2SO_4`

 0,1                      →   0,2(mol)

`CuSO_4+2NaOH→Cu(OH)_2+Na_2SO_4`

  0,3                  →       0,3(mol)

`2Fe(OH)_3`$\xrightarrow[]{t^o}$`Fe_2O_3+3H_2O`

   0,2               →   0,1 (mol)

`Cu(OH)_2`$\xrightarrow[]{t^o}$`CuO+H_2O`

   0,3              → 0,3(mol)

`⇒m_{CR}=0,1.160+0,3.80=40(g)`