câu 1 
\(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\\ pthh:Fe+2HCl\rightarrow FeCl_2+H_2\) 
           0,25   0,5         0,25       0,25 
\(m_{FeCl_2}=0,25.127=31,75g\\ V_{H_2}=0,25.22,4=5,6\\ C_{M\left(HCl\right)}=\dfrac{0,5}{0,2}=2,5M\)
câu 2 
1 ) \(m_{\text{dd}}=35+100=135g\\ 2,C\%=\dfrac{204}{204+100}.100=60\%\\ =>m\text{dd}=\dfrac{100.204}{60}=340g\)
 

`a)PTHH:`

`Zn + 2HCl -> ZnCl_2 + H_2 \uparrow`

`0,2`    `0,4`                                                      `(mol)`

`n_[Zn]=13/65=0,2(mol)`

`b)C_[M_[HCl]]=[0,4]/[0,2]=2(M)`

`c)`

`A + 2HCl -> ACl_2 + H_2 \uparrow`

`0,2`  `0,4`                                                `(mol)`

`=>M_A=[4,8]/[0,2]=24(g//mol)`

     `->A` là `Mg`

\(a)Zn+2HCl\rightarrow ZnCl_2+H_2\\ b)n_{Zn}=\dfrac{13}{65}=0,2mol\\ n_{Zn}=n_{ZnCl_2}=n_{H_2}=0,2mol\\ V_{H_2}=0,2.22,4=4,48l\\ c)n_{HCl}=0,2.2=0,4mol\\ C_{M_{HCl}}=\dfrac{0,4}{0,2}=2M\\ d)m_{ZnCl_2}=0,2.136=27,2g\)

\(n_{H_2}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)

\(a.\)

\(n_{HCl}=2n_{H_2}=0.5\cdot2=1\left(mol\right)\)

\(V_{dd_{HCl}}=\dfrac{1}{2}=0.5\left(M\right)\)

\(b.\)

\(n_X=a\left(mol\right)\)

\(\Rightarrow n_Y=2a\left(mol\right),n_Z=a\left(mol\right),n_T=a\left(mol\right)\)

\(M_X=M\left(\text{g/mol}\right)\)

\(\Rightarrow M_Y=2.7M\left(\text{g/mol}\right),M_Z=\dfrac{7M}{3}\left(\text{g/mol}\right),M_T=\dfrac{347}{60}M\left(\text{g/mol}\right)\)

\(m_{hh}=aM+2a\cdot2.7M+a\cdot\dfrac{7}{3}M+a\cdot\dfrac{347}{60}M=34.7\left(g\right)\)

\(\Rightarrow aM=2.4\)

\(n_{hh}=n_{H_2}=0.5\left(mol\right)\)

\(\Rightarrow a+2a+a+a=0.5\)

\(\Rightarrow a=0.1\)

\(M=\dfrac{2.4}{0.1}=24\left(\text{g/mol}\right)\Rightarrow Mg\)

\(Y=2.7\cdot24=65\left(\text{g/mol}\right)\Rightarrow Zn\)

\(Z=\dfrac{7}{3}\cdot24=56\left(\text{g/mol}\right)\Rightarrow Fe\)

\(T=\dfrac{347}{60}\cdot24=137\left(\text{g/mol}\right)\Rightarrow Ba\)

VddHCl=12=0.5(M)VddHCl=12=0.5(M)

mhh=aM+2a⋅2.7M+a⋅73M+a⋅34760M=34.7(g)mhh=aM+2a⋅2.7M+a⋅73M+a⋅34760M=34.7(g)

nhh=nH2=0.5(mol)nhh=nH2=0.5(mol)

\(n_{Fe}=\dfrac{2,24}{56}=0,04mol\)

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

0,04    0,08      0,04       0,04

\(m_{FeCl_2}=0,04\cdot127=5,08\left(g\right)\)

\(V_{H_2}=0,04\cdot22,4=0,896\left(l\right)\)

\(m_{HCl}=0,08\cdot36,5=2,92\left(g\right)\)

\(m_{ddHCl}=\dfrac{2,92}{5}\cdot100=58,4\left(g\right)\)

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Ta có: \(n_{Fe}=\dfrac{2,24}{56}=0,04\left(mol\right)\)

a. PTHH: Fe + 2HCl ---> FeCl₂ + H₂

b. Theo PT: \(n_{FeCl_2}=n_{H_2}=n_{Fe}=0,04\left(mol\right)\)

=> \(m_{FeCl_2}=0,04.127=5,08\left(g\right)\)

=> \(V_{H_2}=0,04.22,4=0,896\left(lít\right)\)

c. Theo PT: \(n_{HCl}=2.n_{Fe}=2.0,04=0,08\left(mol\right)\)

=> \(m_{HCl}=0,08.36,5=2,92\left(g\right)\)

Ta có: \(C_{\%_{HCl}}=\dfrac{2,92}{m_{dd_{HCl}}}.100\%=5\%\)

=> \(m_{dd_{HCl}}=58,4\left(g\right)\)

a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

b, Ta có: \(n_{Fe}=\dfrac{19,6}{56}=0,35\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Fe}=0,35\left(mol\right)\Rightarrow V_{H_2}=0,35.22,4=7,84\left(l\right)\)

c, \(n_{H_2SO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)

d, \(n_{FeSO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow m_{FeSO_4}=0,35.152=53,2\left(g\right)\)

e, \(C_{M_{FeSO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)

d, \(n_{H_2SO_4}=0,25.1,6=0,4\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{n_{Fe}}{1}< \dfrac{n_{H_2SO_4}}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{Fe}=0,35\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,4-0,35=0,05\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,05.98=4,9\left(g\right)\)