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