a.\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{8,96}{22,4}=0,4mol\)

Gọi \(\left\{{}\begin{matrix}n_{Fe}=x\\n_{Zn}=y\end{matrix}\right.\)

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

 x                                      x ( mol )

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

 y                                        y   ( mol )

Ta có:

\(\left\{{}\begin{matrix}56x+65y=25,55\\x+y=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,35\end{matrix}\right.\)

\(\Rightarrow m_{Fe}=0,05.56=2,8g\)

\(\Rightarrow m_{Zn}=0,35.65=22,75g\)

\(\%m_{Fe}=\dfrac{2,8}{25,55}.100=10,95\%\)

\(\%m_{Zn}=100\%-10,95\%=89,05\%\)

b.\(n_{HCl}=2.0,05+2.0,35=0,8mol\)

\(C_M=\dfrac{n}{V}\Rightarrow V=\dfrac{n}{C_M}=\dfrac{0,8}{2}=0,4l\)

200ml = 0,2l

\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)

Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)

        1         2              1           1

     0,3      0,6             0,3        0,3

a) \(n_{ZnCl2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

\(C_{M_{ZnCl2}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)

b) \(n_{H2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)

c) Pt : \(NaOH+HCl\rightarrow NaCl+H_2O|\)

              1            1              1            1

            0,6          0,6

\(n_{NaOH}=\dfrac{0,6.1}{1}=0,6\left(mol\right)\)

\(m_{NaOH}=0,6.40=24\left(g\right)\)
\(m_{ddNaOH}=\dfrac{24.100}{20}=120\left(g\right)\)

 Chúc bạn học tốt

Gọi \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

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

Theo pt: \(\Rightarrow\left\{{}\begin{matrix}3x+y=0,2\\27x+56y=5,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{19}{470}\\y=\dfrac{37}{470}\end{matrix}\right.\)

\(\%m_{Al}=\dfrac{\dfrac{19}{470}\cdot27}{5,5}\cdot100\%=19,84\%\)

\(\%m_{Fe}=100\%-19,84\%=80,16\%\)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ PTHH:2K+2H_2O\rightarrow2KOH+H_2\uparrow\\ Theo.pt:n_{KOH}=2n_{H_2}=2.0,15=0,3\left(mol\right)\)

\(PTHH:2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O \\ Mol:0,3\rightarrow0,15\\ V_{ddH_2SO_4}=\dfrac{0,15}{2}=0,075\left(l\right)\)

X + nH2O → X(OH)n + n/2 H2.

Ta có: n(H2) = 0,15 mol

→ n(OH-) = 2n (H2) = 0,3 mol

→ n(H+) = 0,3 mol → 2. 2V = 0,3 → V = 0,075 (lít)

Gọi nFe = a (mol); nAl = b (mol)

=> 56a + 27b = 11 (1)

nH2 = 8,96/22,4 = 0,4 (mol)

PTHH: 

Fe + 2HCl -> FeCl2 + H2

a ---> 2a ---> a ---> a

2Al + 6HCl -> 2AlCl3 + 3H2

b ---> 1,5b ---> b ---> b

=> a + 1,5b = 0,4 (2)

Từ (1)(2) => a = 0,1 (mol); b = 0,15 (mol)

mFe = 0,1 . 56 = 5,6 (g)

mAl = 0,2 . 27 = 5,4 (g)

THAM KHẢO :
 

Fe + 2HCl -> FeCl₂ + H₂ (1)

a) 2Al + 6HCl -> 2AlCl₃ + 3H₂ (2)

Gọi khối lượng Fe là x(g) (0<x<11) => n_Fe = x/56 (mol)

Thì m_Al là 11-x(g) => n_Al = (11-x)/27 (mol)

n_H2 = 8,96/22,4 = 0,4 (mol)

Theo PT (1) ta có: n_H2 = n_Fe = x/56 (mol)

Theo PT (2) ta có: n_H2 = 3/2 n_Al = 3/2 . (11-x)/27 = (11-x)/18 (mol)

Theo đề bài, n_H2 thu được là 0,4(mol) nên ta có:

x/56 + (11-x)/18 = 0,4

<=> 18x +56(11-x) = 403,2

<=> x = 5,6 (g)

Do đó: m_Fe = 5,6(g) => n_Fe = 5,6/56 = 0,1 (mol)

m_Al = 11-5,6 = 5,4(g) => n_Al = 5,4/27 = 0,2 (mol)

a) 

\(n_{KClO_3}=\dfrac{24.5}{122.5}=0.2\left(mol\right)\)

\(2KClO_3\underrightarrow{^{^{t^0}}}2KCl+3O_2\)

\(n_{O_2}=\dfrac{3}{2}\cdot0.2=0.3\left(mol\right)\)

\(V_{O_2}=0.3\cdot22.4=6.72\left(l\right)\)

\(n_{O_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

\(2KMnO_4\underrightarrow{^{^{t^0}}}K_2MnO_4+MnO_2+O_2\)

\(0.2...............................................0.1\)

\(n_{KMnO_4\left(bđ\right)}=\dfrac{0.2}{90\%}=\dfrac{2}{9}\left(mol\right)\)

\(m_{KMnO_4}=\dfrac{2}{9}\cdot158=35.11\left(g\right)\)

\(n_{Fe_3O_4}=\dfrac{3.48}{232}=0.015\left(mol\right)\)

\(3Fe+2O_2\underrightarrow{^{^{t^0}}}Fe_3O_4\)

\(0.045....0.03......0.015\)

\(m_{Fe}=0.045\cdot56=2.52\left(g\right)\)

\(V_{O_2}=0.03\cdot22.4=0.672\left(l\right)\)

\(2KClO_3\underrightarrow{^{^{t^0}}}2KCl+3O_2\)

\(0.02......................0.03\)

\(m_{KClO_3}=0.02\cdot122.5=2.45\left(g\right)\)

Tiếp bài của creeper nhé:

c. Ta có: \(n_{ZnO}=\dfrac{4,86}{81}=0,06\left(mol\right)\)

Theo PT₍₁₎: \(n_{HCl}=2.n_{ZnO}=2.0,06=0,12\left(mol\right)\)

Theo PT₍₂₎: \(n_{HCl}=2.n_{Zn}=2.0,1=0,2\left(mol\right)\)

=> \(n_{HCl}=0,12+0,2=0,32\left(mol\right)\)

=> \(m_{HCl}=0,32.36,5=11,68\left(g\right)\)

Ta có: \(C_{\%_{HCl}}=\dfrac{11,68}{m_{dd_{HCl}}}.100\%=12\%\)

=> \(m_{dd_{HCl}}=\dfrac{292}{3}\left(g\right)\)

Theo đề, ta có: 

\(D=\dfrac{\dfrac{292}{3}}{V_{dd_{HCl}}}=1,2\)(g/ml)

=> \(V_{dd_{HCl}}=81,1\left(ml\right)\)