B2:

`1, 2HCl+MgO->H_2O+MgCl_2`

`2, 2NaOH+SO_3->H_2O+Na_2SO_4`

`3, CaO+H_2O->Ca(OH)_2`

`4, Cu +H_2O->Cu(OH)_2`

`5, 2KOH + H_2SO4 -> K_2SO4 + 2H_2O`

`6, 2NaOH +CuCl2`$\buildrel{{t^o}}\over\longrightarrow$`Cu(OH)2 + 2NaCl`

`7, AgNO_3 +HCl->AgCldownarrow+HNO_3`

`8, 2HCl+MgO->H_2O+MgCl_2`

`9, 3H_2O + P_2O_5 -> 2H_3PO_4`

B3:

$\begin{array}{l}a,\ PTHH:\\Mg + 2HCl \to MgC{l_2} + {H_2}\\ MgO + 2HCl ⇒ MgC{l_2} + {H_2}O\\ {n_{{H_2}}} = 0,05mol\\  ⇒ {n_{Mg}} = {n_{{H_2}}} = 0,05mol ⇒ {m_{Mg}} = 1,2g\\  ⇒ {m_{MgO}} = 8g \to {n_{MgO}} = 0,2mol \\ \% {m_{Mg}} = \dfrac{{1,2}}{{9,2}} ·100\%  = 13,04\% \\ \% {m_{MgO}} = 100\%  - 13,04\%  = 86,96\% \\ b,\\ {n_{HCl}} = 2{n_{Mg}} + 2{n_{MgO}} = 0,5mol \to {m_{HCl}} = 18,25g\\  ⇒ m = {m_{{\rm{dd}}HCl}} = 125g\\ c,\\ {n_{MgC{l_2}}} = {n_{Mg}} + {n_{MgO}} = 0,25mol \to {m_{MgC{l_2}}} = 23,75g\\ {m_{{\rm{dd}}}} = {m_{hh}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 134,1g\\ ⇒ C{\% _{MgC{l_2}}} = \dfrac{{23,75}}{{134,1}} · 100\%  = 17,7\%  \end{array}$

1. MgO + 2HCl -> MgCl2 + H2O

2. SO3 + 2NaOH -> Na2SO4 + H2O

3. CaO + H2O -> Ca(OH)2

4. Cu(OH)2 -to-> CuO + H2O

5. 2KOH + H2SO4 -> K2SO4 + 2H2O

6. CuCl2 + Ba(OH)2 -> Cu(OH)2↓ + BaCl2

7. AgNO3 + HCl -> AgCl↓ + HNO3

8. 2HCl + MgO -> MgCl2 + H2O

9. P2O5 + 3H2O -> 2H3PO4