a, \(n_{H_2}=0,25\left(mol\right)\)

Bảo toàn e:

\(2n_{Zn}=2n_{H_2}\Rightarrow n_{Zn}=0,25\left(mol\right)\)

\(\Rightarrow m_{Zn}=16,25\left(g\right)\)

\(\Rightarrow m_{Cu}=13,75\left(g\right)\)

b, \(\%m_{Cu}=\dfrac{13,75}{30}=45,83\%\)

\(\Rightarrow\%m_{Zn}=100\%-45,83\%=54,17\%\)

c, Bảo toàn nguyên tố H:

\(n_{HCl}=2n_{H_2}=0,5\left(mol\right)\)

\(\Rightarrow C\%_{HCl}=\dfrac{0,5.36,5}{200}=9,125\%\)

`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)`

bạn tự làm tiếp ạ

a, Ta có:

\(n_{H2}=\frac{1,344}{22,4}=0,06\left(mol\right)\)

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

0,06____0,06 _____0,06 _____ 0,06

\(m_{FeO}=0,06.56=3,36\left(l\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{FeO}=\frac{3,36}{6,48}.100\%=51,85\%\\\%m_{Ag}=100\%-51,85\%=48,15\%\end{matrix}\right.\)

b, \(n_{NaOH}=0,5.125=0,625\left(mol\right)\)

\(PTHH:2NaOH+FeSO_4\rightarrow Na_2SO_4+Fe\left(OH\right)_2\)

Ban đầu :__ 0,625 ____ 0,06______

Phứng :___0,12_____ 0,06

Sau:____0,505 _____ 0 ______ 0,06 _______0,06

\(\Rightarrow\left\{{}\begin{matrix}CM_{NaOH}=\frac{0,505}{0,5}=1,01M\\CM_{Na2SO4}=\frac{0,06}{0,5}=0,12M\end{matrix}\right.\)

$n_{NO_2}  = 0,13(mol)$

$n_{Zn} = a(mol) ; n_{Al} = b(mol)$

Ta có :

$m_{hh} = 65a + 27b = 2,11(gam)$
Bảo toàn electron : $2a + 3b = 0,13$

Suy ra a = 0,02 ; b = 0,03

$\%m_{Zn} = \dfrac{0,02.65}{2,11}.100\%  = 61,61\%$

$\%m_{Al} = 100\% -61,61\% = 38,39\%$

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(n_{Fe}=n_{H_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

\(\Rightarrow m_{Fe}=0.15\cdot56=8.4\left(g\right)\)

\(m_{Cu}=m_{hh}-m_{Fe}=15-8.4=6.6\left(g\right)\)

\(n_{HCl}=2n_{H_2}=0.15\cdot2=0.3\left(mol\right)\)

\(C_{M_{HCl}}=\dfrac{0.3}{0.2}=1.5\left(M\right)\)