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

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

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

PTHH: Mg + 2HCl → MgCl₂ + H₂

Mol:     0,3      0,6         0,3       0,3

\(m_{Mg}=0,3.24=7,2\left(g\right)\)

b/ \(m_{MgCl_2}=0,3.95=28,5\left(g\right)\)

c/ \(m_{HCl}=0,3.36,5=10,95\left(g\right)\)

a)

$MnO_2 + 4HCl \to MnCl_2 + Cl_2 + 2H_2O$

Theo PTHH :

n Cl2 = n MnO2 = 10,44/87 = 0,12(mol)

=> V Cl2 = 0,12.22,4 = 2,688(lít)

b)

$2NaOH + Cl_2 \to NaCl + NaClO + H_2O$

n NaOH = 2n Cl2 = 0,24(mol)

=> V dd NaOH = 0,24/2 = 0,12(lít)

400ml = 0,4l

\(n_{NaOH}=2.0,4=0,8\left(mol\right)\)

a) Pt : \(SO_2+2NaOH\rightarrow Na_2SO_3+H_2O|\)

            1             2                  1               1

          0,4          0,8               0,4

b) \(n_{SO2}=\dfrac{0,8.1}{2}=0,4\left(mol\right)\)

\(V_{SO2\left(dktc\right)}=0,4.22,4=8,96\left(l\right)\)

c) \(n_{Na2SO3}=\dfrac{0,8.1}{2}=0,4\left(mol\right)\)

⇒ \(m_{Na2SO3}=0,4.126=50,4\left(g\right)\)

c) \(C_{M_{Na2SO3}}=\dfrac{0,4}{0,4}=1\left(M\right)\)

 Chúc bạn học tốt

Học tốt nheee

a, \(CO_2+2KOH\rightarrow K_2CO_3+H_2O\)

b, \(n_{KOH}=3.2=6\left(mol\right)\)

Theo PT: \(n_{CO_2}=n_{K_2CO_3}=\dfrac{1}{2}n_{KOH}=3\left(mol\right)\)

\(\Rightarrow V_{CO_2}=3.24,79=74,37\left(l\right)\)

c, \(C_{M_{K_2CO_3}}=\dfrac{3}{3}=1\left(M\right)\)

a) Gọi $n_{Al} = a(mol) ; n_{Fe} = b(mol) \Rightarrow 27a + 56b = 33,4(1)$

$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$Fe + 2HCl \to FeCl_2 + H_2$
Theo PTHH : $n_{H_2} = 1,5a + b = \dfrac{17,92}{22,4} = 0,8(2)$

Từ (1)(2) suy ra : a = 0,2 ; b = 0,5

$\%m_{Al} = \dfrac{0,2.27}{33,4}.100\% = 16,17\%$
$\%m_{Fe} = 100\% - 16,17\% = 83,83\%$

b) $n_{HCl} = 2n_{H_2} = 1,6(mol)$

c) $m_{muối} = m_{hh} + m_{HCl} - m_{H_2} = 33,4 + 1,6.36,5 - 0,8.2 = 90,2(gam)$