Bài 8:

\(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)

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

_____0,2______0,6_____0,2____0,3 (mol)

a, \(m_{Al}=0,2.27=5,4\left(g\right)\)

b, \(C_{M_{HCl}}=\dfrac{0,6}{0,3}=2\left(M\right)\)

c, \(C_{M_{AlCl_3}}=\dfrac{0,2}{0,2}=1\left(M\right)\)

Bài 9:

Ta có: \(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

a, \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right)\)

\(\Rightarrow m_{MgO}=8,4-2,4=6\left(g\right)\)

b, \(n_{MgO}=\dfrac{6}{40}=0,15\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Mg}+2n_{MgO}=0,5\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,5.36,5}{3,65\%}==500\left(g\right)\)

nHCl = 200.0,073/36,5 = 0,4(mol)
2M + 2nHCl → 2MCln + nH2
0,4/n _ 0,4 ___ 0,4/n __ 0,2

mddMCln = 200 + M.0,4/n - 0,2.2 = 199,6 + 0,4M/n
Nồng độ muối MCln bằng 11,96%
⇒ (M + 35,5n).0,4/n = 0,1196.(199,6 + 0,4M/n)
⇒ M = 27,5n
n = 2, M = 55 (Mn)
x = 55.0,4/2 = 11(g)

A. Fe3O4.

\(n_{HCl}=\dfrac{200\cdot7.3\%}{36.5}=0.4\left(mol\right)\)

\(M+2HCl\rightarrow MCl_2+H_2\)

\(0.2.....0.4.........0.2........0.2\)

\(m_{MCl_2}=0.2\cdot\left(M+71\right)\left(g\right)\)

\(m_{dd}=0.2M+200-0.2\cdot2=0.2M+199.6\left(g\right)\)

\(C\%MCl_2=\dfrac{0.2\cdot\left(M+71\right)}{0.2M+199.6}\cdot100\%=12.05\%\)

\(\Rightarrow M=56\)

\(M:Sắt\)

\(a.PTHH:Zn+2HCl\xrightarrow[]{}ZnCl_2+H_2\\ b.n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ PTHH:Zn+2HCl\xrightarrow[]{}ZnCl_2+H_2\\ n_{H_2}=0,2.2=0,4\left(mol\right)\\ V_{H_2}=0,4.22,4=8,96\left(l\right)\\ c.n_{HCl}=n_{Zn}=0,2mol\\ C_{MHCl}=\dfrac{0,4}{0,1}=4\left(M\right)\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH :

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

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

\(b,V_{H_2}=0,2.22,4=4,48\left(l\right)\)

\(c,C_M=\dfrac{n}{V}=\dfrac{0,4}{0,1}=4M\)

\(M + 2HCl \to MCl_2 + H_2\\ n_M = n_{H_2} = \dfrac{2,24}{22,4} = 0,1(mol)\\ \Rightarrow M = \dfrac{5,6}{0,1} = 56(Fe)\\ \)

Vậy M là kim loại Fe

\(n_{FeCl_2} = n_{H_2} = 0,1(mol)\\ m_{FeCl_2} = 0,1.127 = 12,7(gam)\\ m_{dd\ sau\ pư} =m_{Fe} + m_{dd\ HCl} -m_{H_2} = 5,6 + 94,6 -0,1.2 = 100(gam)\\ C\%_{FeCl_2} = \dfrac{12,7}{100}.100\% = 12,7\%\)

Giả sử có 1 mol RCO₃

PTHH: RCO₃ + 2HCl → RCl₂ + CO₂↑ + H₂O

                1      →   2           1          1           1 (mol)

Ta có m_{dd(sau pứ)} = mRCO3+mdd(HCl)−mCO2

⟹ m_{dd(sau pứ)} = \(\dfrac{\text{MR+60+(2.36,5.100%)}}{7,3\%-1,44}\)=MR+1016

⟹ C%(RCl2)=mRCl2mdd(sau).100%=\(\dfrac{M_{R_{ }}+71}{_{ }M_R+1016}\).100%=9,135%

⟹ M_R = 24 (Mg)

Vậy công thức của muối là MgCO₃.