a, \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

BT e, có: 3n_Fe = 2n_SO2 ⇒ n_SO2 = 0,45 (mol)

⇒ V_SO2 = 0,45.22,4 = 10,08 (l)

b, BTNT Fe, có: n_Fe(OH)3 = n_Fe = 0,3 (mol)

⇒ m_Fe(OH)3 = 0,3.107 = 32,1 (g)

a) n Fe = a(mol) ; n Cu = b(mol)

=> 56a + 64b = 4,8(1)

n SO2 = 2,24/22,4 = 0,1(mol)

Bảo toàn e :

3n Fe + 2n Cu = 2n SO2

<=> 3a + 2b = 0,2(2)

Từ (1)(2) suy ra a = b = 0,04

%m Fe = 0,04.56/4,8  .100% = 46,67%

%m Cu = 100% -46,67% = 53,33%

b)

n KOH = 0,15

Ta có :

1 < n KOH / n SO2 = 0,15/0,1 = 1,5 < 2 nên muối sinh ra là Na2SO3(x mol) và NaHSO3(y mol)

2NaOH + SO2 $\to$ Na2SO3 + H2O

NaOH + SO2 $\to$ NaHSO3

Theo PTHH :

n SO2 = a + b = 0,1

n NaOH = 2a + b = 0,15

Suy ra a= 0,05 ; b = 0,05

m muối = 0,05.126 + 0,05.104 = 11,5 gam

Bài 2 :

- Gọi số mol Fe và Cu lần lượt là a, b mol

Ta có : mhh = mFe + mCu = 56a + 64b = 4,8

Bte : 3a + 2b = 0,2 

=> a = b =0,04mol

a, Ta có : mFe =n.M = 2,24g ( 46,7% )

=> %Cu = 53,3%

b, Ta thấy sau phản ứng thu được K+, SO3-2 xmol, HSO3- y mol

BtS : x + y = 0,1 

BTĐT : 2x + y = 0,15 

=> x =y = 0,05 mol

=> mM = mK2SO3 + mKHSO3 = 13,9g

\(n_{Fe}=a\left(mol\right),n_{Zn}=b\left(mol\right)\)

\(m=56a+65b=13.22\left(g\right)\left(1\right)\)

\(n_{H_2}=\dfrac{4.928}{22.4}=0.22\left(mol\right)\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(n_{H_2}=a+b=0.22\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):\)

\(a=0.12\)

\(b=0.1\)

\(\text{Bảo toàn e : }\)

\(n_{Zn}+n_{Fe}=n_{SO_2}=\dfrac{0.12}{2}+\dfrac{0.1}{2}=0.11\left(mol\right)\)

\(V_{SO_2}=0.11\cdot22.4=2.464\left(l\right)\)

`2Fe + 6H_2 SO_[4(đ,n)] -> Fe_2(SO_4)_3 + 3SO_2 \uparrow + 6H_2 O`

`0,05`        `0,15`                               `0,025`                                     `(mol)`

`Cu + 2H_2 SO_[4(đ,n)] -> CuSO_4 + SO_2 \uparrow + 2H_2 O`

`0,225`     `0,45`                         `0,225`                                          `(mol)`

`n_[SO_2]=[6,72]/[22,4]=0,3(mol)`

Gọi `n_[Fe]=x` ; `n_[Cu]=y`

`=>` $\begin{cases} \dfrac{3}{2}x+y=0,3\\56x+64y=17,2 \end{cases}$

`<=>` $\begin{cases}x=0,05\\y=0,225 \end{cases}$

  `@m_[Fe_2(SO_4)_3]=0,025.400=10(g)`

  `@m_[CuSO_4]=0,225.160=36(g)`

  `@m_[dd H_2 SO_4]=[(0,15+0,45).98]/80 .100=73,5(g)`

Sửa đề: 80% ---> 98% (80% chưa đặc nên không giải phóng SO₂ được)

Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Cu}=b\left(mol\right)\end{matrix}\right.\)

\(\rightarrow56a+64b=17,2\left(1\right)\)

PTHH: 

\(2Fe+6H_2SO_{4\left(đặc,nóng\right)}\rightarrow Fe_2\left(SO_4\right)_3+3SO_2\uparrow+6H_2O\)

a------>3a------------------->0,5a--------------->1,5a

\(Cu+2H_2SO_{4\left(đặc,nóng\right)}\rightarrow CuSO_4+SO_2\uparrow+2H_2O\)

b----->2b------------------->b------------->b

\(\rightarrow1,5a+b=\dfrac{6,72}{22,4}=0,3\left(2\right)\)

Từ \(\left(1\right)\left(2\right)\rightarrow\left\{{}\begin{matrix}a=0,05\left(mol\right)\\b=0,225\left(mol\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{Fe_2\left(SO_4\right)_3}=0,5.0,05.400=10\left(g\right)\\m_{CuSO_4}=0,225.160=36\left(g\right)\\m_{ddH_2SO_4}=\dfrac{\left(0,05.3+0,225.2\right).98}{98\%}=60\left(g\right)\end{matrix}\right.\)

`2Fe + 6H_2 SO_[4(đ,n)] -> Fe_2(SO_4)_3 + 6H_2 O + 3SO_2 \uparrow`

 `0,1`                                                                                `0,15`        `(mol)`

`2Ag + 2H_2 SO_[4(đ,n)] -> Ag_2 SO_4 + 2H_2 O + SO_2 \uparrow`

`0,2`                                                                         `0,1`               `(mol)`

`n_[SO_2]=[5,6]/[22,4]=0,25(mol)`

Gọi `n_[Fe]=x` ; `n_[Ag]=y`

`=>` $\left[\begin{matrix} 56x+108y=27,2\\ \dfrac{3}{2}x+\dfrac{1}{2}y=0,25\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=0,1\\ y=0,2\end{matrix}\right.$

`a)%m_[Fe]=[0,1.56]/[27,2] .100~~20,59%`

`=>%m_[Ag]~~100-20,59~~79,41%`

`b)n_[SO_2]=0,15+0,1=0,25(mol)`

   `n_[NaOH]=0,5.0,8=0,4(mol)`

Ta có:`T=[0,4]/[0,25]=1,6 ->` Tạo muối `Na_2 SO_3` và `NaHSO_3`

 `SO_2 + 2NaOH -> Na_2 SO_3 + H_2 O`

 `SO_2 + NaOH -> NaHSO_3`

Gọi `n_[Na_2 SO_3]=x ; n_[NaHSO_3]=y`

 `=>` $\left[\begin{matrix} x+y=0,25\\ 2x+y=0,4\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=0,15\\ y=0,1\end{matrix}\right.$

    `=>C_[M_[Na_2 SO_3]]=[0,15]/[0,5]=0,3(M)`

    `=>C_[M_[NaHSO_3]]=[0,1]/[0,5]=0,2(M)`

a.b.

\(n_{Al}=\dfrac{4,05}{27}=0,15mol\)

\(2Al+6H_2SO_4\left(đ\right)\rightarrow\left(t^o\right)Al_2\left(SO_4\right)_3+3SO_2+6H_2O\)

0,15                                                        0,225                 ( mol )

\(V_{SO_2}=0,225.22,4=5,04l\)

c.

\(Ba\left(OH\right)_2+SO_2\rightarrow BaSO_3+H_2O\)

\(Ba\left(OH\right)_2+2SO_2\rightarrow Ba\left(HSO_3\right)_2\)

Gọi \(\left\{{}\begin{matrix}n_{BaSO_3}=x\\n_{Ba\left(HSO_3\right)_2}=y\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}217x+299y=38,7\\x+2y=0,225\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,075\\y=0,075\end{matrix}\right.\)

\(n_{Ba\left(OH\right)_2}=0,075+0,075=0,15mol\)

\(V_{Ba\left(OH\right)_2}=\dfrac{0,15}{1}=0,15l\)

a.b.

VBa(OH)2=0,151=0,15l