a, 

2Na+2H2O->2NaOH+H2

NaOH+Ca(HCO3)2->NaHCO3+CaCO3+H2O

NaHCO3+NaOH->Na2CO3+H2O

Na2CO3+NaHSO4->NaHCO3+Na2SO4

Na2O+H2O->2NaOH

Na2CO3+Ca(OH)2->2NaOH+CaCO3

a,

2Na + 2H2O→2NaOH +H2

4Na + O2 → 2Na2O

2Na2O +  H2O  → 2NaOH

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ CuSO_4+Ba\left(NO_3\right)_2\rightarrow BaSO_4+Cu\left(NO_3\right)_2\\ Cu\left(NO_3\right)_2+2KOH\rightarrow Cu\left(OH\right)_2+2KNO_3\\ Cu\left(OH\right)_2\rightarrow\left(t^o\right)CuO+H_2O\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ CuCl_2\rightarrow\left(t^o\right)Cu+Cl_2\)

43.a)  \(m_{HCl\left(bđ\right)}=200.10,95\%=21,9\left(g\right)\)

=> \(n_{HCl\left(bđ\right)}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)

b) HCl phản ứng với NaOH là HCl dư

 \(HCl+NaOH\rightarrow NaCl+H_2O\)

\(n_{HCl\left(dư\right)}=n_{NaOH}=0,05.2=0,1\left(mol\right)\)

=> \(n_{HCl\left(pứ\right)}=n_{HCl\left(bđ\right)}-n_{HCl\left(dư\right)}=0,6-0,1=0,5\left(mol\right)\)

c) \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)

\(n_{CaCO_3}=\dfrac{1}{2}n_{HCl\left(pứ\right)}=0,25\left(mol\right)\)

=> \(m_{CaCO_3}=0,25.100=25\left(g\right)\)

d) \(n_{CO_2}=\dfrac{1}{2}n_{HCl\left(pứ\right)}=0,25\left(mol\right)\)

=> \(V_{CO_2}=0,25.22,4=5,6\left(l\right)\)

e) \(m_{ddsaupu}=25+200-0,25.44=214\left(g\right)\)

Dung dịch A gồm CaCl2 và HCl dư

\(n_{CaCl_2}=\dfrac{1}{2}n_{HCl\left(pứ\right)}=0,25\left(mol\right)\)

\(C\%_{CaCl_2}=\dfrac{0,25.111}{214}.100=12,97\%\)

\(C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{214}.100=1,71\%\)

a, \(n_{CO_2}=\dfrac{17,6}{44}=0,4\left(mol\right)\)

CH₄ + 2O₂ -----t^o---> CO₂ + 2H₂O

 x                                 x

C₂H₄ + 3O₂ -----t^o---> 2CO₂ + 2H₂O

 y                                    2y

Ta có hệ pt: \(\left\{{}\begin{matrix}16x+28y=6\\x+2y=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

\(\%m_{CH_4}=\dfrac{0,2.16.100\%}{6}=53,33\%;\%m_{C_2H_4}=100\%-53,33\%=46,67\%\)

b, \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,2.22,4.100\%}{\left(0,2+0,1\right).22,4}=66,67\%\\\%V_{C_2H_4}=100\%-66,67\%=33,33\%\end{matrix}\right.\)

\(BaCl_2+K_2SO_4\rightarrow BaSO_4+2KCl\)

a) (1) 2Fe + 3Cl₂ ➙ 2FeCl₃

(2) FeCl₃ + 3NaOH ➙ Fe(OH)₃ + 3NaCl

(3) 2Fe(OH)₃ ➙ Fe₂O₃ + 3H₂O

(4)Fe₂O₃ + 3H₂SO₄ ➙ Fe₂(SO₄)₃ + 3H₂O

(5) Fe₂(SO₄)₃ + 3BaCl₂ ➙ 2FeCl₃ + 3BaSO₄
b) (1) 2Al₂O₃ ➙ 4Al + 3O₂

(2)2Al + 6HCl ➙ 2AlCl₃ + 3H₂

(3) AlCl₃ + 3NaOH ➙ Al(OH)₃ + 3NaCl

(4) 2Al(OH)₃ ➙ Al₂O₃ + 3H₂O

(5) Al₂O₃ + 6HCl ➙ 2AlCl₃ + 3H₂O

4 cách điều chế NaOH :

Pt : \(2Na+2H_2O\rightarrow2NaOH+H_2\)

       \(Na_2O+H_2O\rightarrow2NaOH\)

        \(2NaCl+H_2O\xrightarrow[có.màng.ngăn]{điện.phân}2NaOH+H_2+Cl_2\)

         \(Ba\left(OH\right)_2+Na_2SO_4\rightarrow BaSO_4+2NaOH\)

  Chúc bạn học tốt

Mik viết PTHH nhé:

\(-2NaCl+2H_2O\xrightarrow[có.màng.ngăn]{điện.phân}Cl_2+H_2+2NaOH\)

\(-2Na+2H_2O--->2NaOH+H_2\)

\(-Na_2O+H_2O--->2NaOH\)

\(-Ca\left(OH\right)_2+2NaHCO_3--->CaCO_3\downarrow+2NaOH+H_2O\)

bài 28 ý hả bn! bn nêu đề ra nha!

mk dùng sách vnen nên ko có mấy cái đó.