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

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

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

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

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

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

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

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

.

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

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

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

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