a ) PTHH : \(Na_2O+H_2O\rightarrow2NaOH\)

b ) \(PT:Na_2O+H_2O\rightarrow2NaOH\)

                \(0,25\)                      \(0,5\)    ( mol )

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

                 \(0,5\)          \(0,5\)                   ( mol )

\(V_{HCl}=\frac{0,5}{1}=0,5\left(lít\right)\)

\(n_{HCl}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PTHH: NaOH + HCl --> NaCl + H₂O

_______0,6<--0,6

=> mNaOH = 0,6.40 = 24(g)

=> \(m_{dd}=\dfrac{24.100}{10}=240\left(g\right)\)

a, \(Na_2O+H_2O\rightarrow2NaOH\)

\(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)

Theo PT: \(n_{NaOH}=0,25.2=0,5\left(mol\right)\)

\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,5}{0,5}=1\left(M\right)\)

b, \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,25.98=24,5\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{24,5}{20\%}=122,5\left(g\right)\)

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,5\left(ml\right)\)

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

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

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

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

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

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

 Chúc bạn học tốt

Lần sau bạn đăng tách từng bài ra nhé.

Câu 1:

a, \(Na_2O+H_2O\rightarrow2NaOH\)

\(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)

Theo PT: \(n_{NaOH}=2n_{Na_2O}=0,5\left(mol\right)\Rightarrow C_{M_{NaOH}}=\dfrac{0,5}{0,5}=1\left(M\right)\)

b, \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,25.98=24,5\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{24,5}{20\%}=122,5\left(g\right)\)

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,46\left(ml\right)\)

Câu 3: \(n_{CO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)

\(n_{CaCO_3}=n_{CO_2}=0,4\left(mol\right)\Rightarrow m_{CaCO_3}=0,4.100=40\left(g\right)\)

Câu 4: \(n_{CuSO_4}=\dfrac{32}{160}=0,2\left(mol\right)\)

\(n_{BaCl_2}=\dfrac{20,8}{208}=0,1\left(mol\right)\)

PT: \(CuSO_4+BaCl_2\rightarrow BaSO_{4\downarrow}+CuCl_2\)

Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,1}{1}\), ta được CuSO₄ dư.

Theo PT: \(n_{BaSO_4}=n_{BaCl_2}=0,1\left(mol\right)\Rightarrow m_{BaSO_4}=0,1.233=23,3\left(g\right)\)

Câu 2:

\(n_{HCl}=0,3.0,2=0,06\left(mol\right)\)

PT: \(HCl+NaOH\rightarrow NaCl+H_2O\)

Theo PT: \(n_{NaOH}=n_{NaCl}=n_{HCl}=0,06\left(mol\right)\)

\(\Rightarrow V_{NaOH}=\dfrac{0,06}{1,5}=0,04\left(l\right)\)

\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,06}{0,3+0,04}=\dfrac{3}{17}\left(M\right)\)

\(a,PTHH:Na_2O+H_2O\rightarrow2NaOH\\ \Rightarrow n_{NaOH}=2n_{Na_2O}=2\cdot\dfrac{37,2}{62}=0,6\cdot2=1,2\left(mol\right)\\ \Rightarrow C_{M_{NaOH}}=\dfrac{1,2}{0,5}=2,4M\\ b,PTHH:2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\\ \Rightarrow n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,6\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=0,6\cdot98=58,8\left(g\right)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{58,8\cdot100\%}{20\%}=294\left(g\right)\\ \Rightarrow V_{dd}=\dfrac{294}{1,14}\approx257,9\left(ml\right)\)

đề bài ko rõ lắm, 0,51 hay là sao

\(n_{Na_2O}=\dfrac{15,5}{62}=0,25mol\)

\(Na_2O+H_2O\rightarrow2NaOH\)

0,25                       0,5

\(C_{M_{NaOH}}=\dfrac{0,5}{0,51}=0,98M\)

Để trung hòa: \(n_{OH^-}=n_{H^+}=0,5mol\)

\(\Rightarrow n_{H_2SO_4}=\dfrac{1}{2}n_{H^+}=\dfrac{1}{2}\cdot0,5=0,25mol\)

PT:Na2O+H2O --->2NaOH

nNa2O= 15,5/62=0,25(mol) 

Theo PT, ta có: nNaOH=1/2nNa2O=0,125 mol

=>CM dd NaOH=0,125/0,5=0,25M

Ta có: \(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)

PTHH: Na₂O + H₂O ---> 2NaOH

Theo PT: \(n_{NaOH}=\dfrac{1}{2}.n_{Na_2O}=\dfrac{1}{2}.0,25=0,125\left(mol\right)\)

=> \(C_{M_{NaOH}}=\dfrac{n_{NaOH}}{V_{dd_{NaOH}}}=\dfrac{0,125}{0,5}=0,25\)(g/mol)