Fe+2HCl->FeCl₂+H₂

0,125--0,25---0,125-0,125

m _HCl=9,125 g=>n _HCl=\(\dfrac{9,125}{26,5}\)=0,25 mol

=>m Fe=0,125.56=7g

=>V_H2=0,125.22,4=2,8l

=>C%_FeCl2=\(\dfrac{0,125.127}{7+182,5-0,25}\).100=8,388%

 \(Fe+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Fe+H_2\)

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

Theo PT: \(n_{\left(CH_3COO\right)_2Fe}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{\left(CH_3COO\right)_2Fe}=0,2.174=34,8\left(g\right)\)

Ta có: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\)

m _{dd sau pư} = 11,2 + 200 - 0,2.2 = 210,8 (g)

\(\Rightarrow C\%_{\left(CH_3COO\right)_2Fe}=\dfrac{34,8}{210,8}.100\%\approx16,51\%\)

\(a) Fe + 2HCl \to FeCl_2\\ b) n_{HCl} = \dfrac{182,5.5\%}{36,5} = 0,25(mol)\\ n_{FeCl_2} = n_{H_2} = n_{Fe} = \dfrac{1}{2}n_{HCl} = 0,125(mol)\\ \Rightarrow m_{Fe} = 0,125.56 = 7(gam) ; V = 0,125.22,4 = 2,8(lít)\\ c) m_{dd\ sau\ phản\ ứng} = m_{Fe} + m_{dd\ HCl} - m_{H_2} = 7 + 182,5 - 0,125.2 = 189,25(gam)\\ C\%_{FeCl_2} = \dfrac{0,125.127}{189,25}.100\% = 8,39\%\)

\(n_{H_2SO_4}=0.3\cdot1=0.3\left(mol\right)\)

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

\(0.3.....0.3..............0.3...........0.3\)

\(m_{Fe}=0.3\cdot56=16.8\left(g\right)\)

\(V_{H_2}=0.3\cdot22.4=6.72\left(l\right)\)

\(C_{M_{FeSO_4}}=\dfrac{0.3}{0.3}=1\left(M\right)\)

200ml = 0,2l

\(n_{CuSO4}=1.0,2=0,2\left(mol\right)\)

a) Pt : \(CuSO_4+2KOH\rightarrow Cu\left(OH\right)_2+K_2SO_4|\)

               1               2                1                  1

              0,2          0,4               0,2                0,2

b) \(n_{KOH}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\) 

200ml = 0,2l

\(C_{M_{ddKOH}}=\dfrac{0,4}{0,2}=2\left(M\right)\)

c) \(n_{Cu\left(OH\right)2}=\dfrac{0,4.1}{2}=0,2\left(mol\right)\)

⇒ \(m_{Cu\left(OH\right)2}=0,2.98=19,6\left(g\right)\)

d) \(n_{K2SO4}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(V_{ddspu}=0,2+0,2=0,4\left(l\right)\)

\(C_{M_{K2SO4}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)

 Chúc bạn học tốt

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)  
          0,2       0,4           0,2 
\(m_{MgCl_2}=0,2.95=19\left(g\right)\\ C\%_{HCl}=\dfrac{0,4.36,5}{200}.100\%=7,3\%\)

\(n_{H_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)

a)

PTHH:

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

0,15<-0,3<-------------0,15

b)

\(m_{Fe}=0,15.56=8,4\left(g\right)\)

c)

\(CM_{HCl}=\dfrac{0,3}{0,05}=6M\)

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

b, Ta có: \(n_{H_2SO_4}=0,5.1=0,5\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{H_2SO_4}=0,5\left(mol\right)\Rightarrow m_{Fe}=0,5.56=28\left(g\right)\)

c, \(n_{FeSO_4}=n_{H_2SO_4}=0,5\left(mol\right)\Rightarrow C_{M_{FeSO_4}}=\dfrac{0,5}{0,5}=1\left(M\right)\)

\(a)Fe+CuSO_4\rightarrow FeSO_4+Cu\\ FeSO_4+2NaOH\rightarrow Fe\left(OH\right)_2+Na_2SO_4\\ b)n_{NaOH}=0,2.1=0,2mol\\ n_{Fe}=n_{Fe\left(OH\right)_2}=0,2:2=0,1mol\\ m_{Fe}=0,1.56=5,6g\)