pứ: Fe + 2HCl -> FeCl₂ + H₂

b. n_Fe = \(\dfrac{5,6}{56}\)= 0,1 mol

Từ pt suy ra được: n_HCl = 2.n_Fe= 0,2 mol

=> m_HCl = 0,2. 36,5 = 7,3 g

c. n_H2 = n_Fe = 0,1 mol

=> V_H2 = 0,1.22,4 = 2,24 (lít)

\(a)n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ n_{HCl}=0,5.1=0,5mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{0,5}{2}\Rightarrow HCl.dư\\ Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1         0,2          0,1          0,1

\(m_{HCl\left(dư\right)}=\left(0,5-0,2\right).36,5=10,95g\\ b)m_{FeCl_2}=0,1.127=12,7g\\ c)V_{H_2}=0,1.22,4=2,24l\\ d)C_{\%HCl\left(dư\right)}=\dfrac{10,95}{200}\cdot100=5,475\%\\ C_{\%HCl\left(pư\right)}=\dfrac{0,2.36,5}{200}\cdot100=3,65\%\)

Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

____0,1______0,2_____0,1____0,1 (mol)

a, \(C_{M_{HCl}}=\dfrac{0,2}{0,15}=\dfrac{4}{3}\left(M\right)\)

\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b, \(FeCl_2+2NaOH\rightarrow2NaCl+Fe\left(OH\right)_2\)

Theo PT: \(n_{NaOH}=2n_{FeCl_2}=0,2\left(mol\right)\)

\(\Rightarrow V_{NaOH}=\dfrac{0,2}{2}=0,1\left(l\right)\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1mol\)

Fe + 2HCl \(\rightarrow\) FeCl₂ + H₂

0,1                  0,1                ( mol )

\(m_{FeCl_2}=0,1.127=12,7g\)          

n_H2= \(\dfrac{6,72}{22,4}\)= 0,3 ( mol )

Fe + 2HCl ---> FeCl₂ + H₂ \(\uparrow\)

0,3 0,6 0,3 0,3 ( mol )

a. m_Fe= 0,3 . 56 = 16,8 (g)

b. V_HCl= 0,6 .22,4 = 13,44 (l)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

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

Theo PT: \(\left\{{}\begin{matrix}n_{H_2}=n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\\n_{HCl}=2n_{Fe}=0,4\left(mol\right)\end{matrix}\right.\)

a, Ta có: \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

b, \(V_{ddHCl}=\dfrac{0,4}{1,5}\approx0,267\left(l\right)\)

c, \(m_{FeCl_2}=0,2.127=25,4\left(g\right)\)

Bạn tham khảo nhé!