\(n_{Na}=\dfrac{2.3}{23}=0.1\left(mol\right)\)

\(n_{HCl}=0.4\cdot2=0.8\left(mol\right)\)

\(NaOH+HCl\rightarrow NaCl+H_2O\)

\(0.8..............0.8\)

\(2Na+2H_2O\rightarrow2NaOH+H_{_{ }2}\)

\(0.1........................0.1\)

\(n_{NaOH}=0.1< 0.8\)

Đề nhầm lãn !

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\(n_K=\dfrac{7,8}{39}=0,2\left(mol\right)\)

Pt : \(2K+2H_2O\rightarrow2KOH+H_2|\)

       2           2               2          1

     0,2                         0,2

a) \(n_{KOH}=\dfrac{0,2.2}{2}=0,2\left(mol\right)\)

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

b) Pt : \(HCl+KOH\rightarrow KCl+H_2O|\)

             1            1            1          1

           0,2          0,2

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

\(m_{HCl}=0,2.36,5=7,3\left(g\right)\)

\(m_{ddHCl}=\dfrac{7,3.100}{15}\simeq48,67\left(g\right)\)

\(V_{ddHCl}=\dfrac{48,67}{1,2}=40,56\left(ml\right)\)

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\(n_{Na_2O}=\dfrac{12,4}{62}=0,2mol\)

\(Na_2O+H_2O\rightarrow2NaOH\)

0,2  \(\rightarrow\)    0,2   \(\rightarrow\)   0,4

\(C_{M_{NaOH}}=\dfrac{0,4}{\dfrac{500}{1000}}=0,8M\)

PTHH:SO3+H2O--->H2SO4

Ta có

n\(_{SO3}=\frac{16}{80}=0,2\left(mol\right)\)

Theo pthh

n\(_{H2SO4}=n_{SO3}=0,2\left(mol\right)\)

C\(_{M\left(H2SO4\right)}=\frac{0,2}{0,4}=0,5\left(M\right)\)

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nSO3 = 0,2 mol

PTHH : SO3 +H2O--->H2SO4

=>nH2SO4 = 0,2 mol ( do nH2SO4 = nSO3)

=>CM = 0,2 / 0,4 = 0,5 M

Câu 1 : 

\(n_{K_2O}=\dfrac{2.35}{94}=0.025\left(mol\right)\)

\(K_2O+H_2O\rightarrow2KOH\)

\(0.025...................0.05\)

\(C_{M_{KOH}}=\dfrac{0.05}{0.4}=0.125\left(M\right)\)

Câu 2 : 

\(n_{Ca\left(OH\right)_2}=\dfrac{1.11}{111}=0.01\left(mol\right)\)

\(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)

\(0.01..........................0.01\)

\(C_{M_{CaCl_2}}=\dfrac{0.01}{0.5}=0.02\left(M\right)\)

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

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