$n_{OH^-} = 0,12.1,2+0,1.0,12.2 = 0,168$ mol ; $n_{Ba^{2+}} = 0,012$ mol $n_{SO_4^{2-}} = 0,02$ mol => $n_{BaSO_4} = n_{Ba^{2+}} = 0,012$ mol ; $m_{BaSO_4} = 2,796 < 3,732 $=> Có kết tủa $Al(OH)_3$ ; $m_{AlOH)_3} = 3,732-2,796 = 0,936$ gam => $n_{AlOH)_3} = 0,012$ mol$H^+ + OH^- => H_2O$$0,1 0,1$$Al^{3+} + 3OH^- => AlOH)_3$$z 3z z$$Al(OH)_3 + OH^- => AlO_2^- + 2H_2O$$a a$Theo đề ta có $\left\{ \begin{array}{l} 0,1+3z+a=0,168\\ 3z-a=0,012 \end{array} \right. => \left\{ \begin{array}{l} a=0,008\\ z=0,02\end{array} \right.$Theo ĐLBT điện tich : $n_{NO_3^-} =t= 3z+0,1-0,02.2 = 0,12$ mol
$n_{OH^-} = 0,12.1,2+0,1.0,12.2 = 0,168$ mol ; $n_{Ba^{2+}} = 0,012$ mol $n_{SO_4^{2-}} = 0,02$ mol => $n_{BaSO_4} = n_{Ba^{2+}} = 0,012$ mol ; $m_{BaSO_4} = 2,796 < 3,732 $=> Có kết tủa $Al(OH)_3$ ; $m_{AlOH)_3} = 3,732-2,796 = 0,936$ gam => $n_{AlOH)_3} = 0,012$ mol$H^+ + OH^- => H_2O$$0,1 0,1$$Al^{3+} + 3OH^- => AlOH)_3$$z 3z 3z$$Al(OH)_3 + OH^- => AlO_2^- + 2H_2O$$a a$Theo đề ta có $\left\{ \begin{array}{l} 0,1+3z+a=0,168\\ 3z-a=0,012 \end{array} \right. => \left\{ \begin{array}{l} a=0,028\\ z=1/75\end{array} \right.$Theo ĐLBT điện tich : $n_{NO_3^-} =t= 3z+0,1-0,02.2 = 0,1$ mol
$n_{OH^-} = 0,12.1,2+0,1.0,12.2 = 0,168$ mol ; $n_{Ba^{2+}} = 0,012$ mol $n_{SO_4^{2-}} = 0,02$ mol => $n_{BaSO_4} = n_{Ba^{2+}} = 0,012$ mol ; $m_{BaSO_4} = 2,796 < 3,732 $=> Có kết tủa $Al(OH)_3$ ; $m_{AlOH)_3} = 3,732-2,796 = 0,936$ gam => $n_{AlOH)_3} = 0,012$ mol$H^+ + OH^- => H_2O$$0,1 0,1$$Al^{3+} + 3OH^- => AlOH)_3$$z 3z z$$Al(OH)_3 + OH^- => AlO_2^- + 2H_2O$$a a$Theo đề ta có $\left\{ \begin{array}{l} 0,1+3z+a=0,168\\ 3z-a=0,012 \end{array} \right. => \left\{ \begin{array}{l} a=0,0
08\\ z=
0,02\end{array} \right.$Theo ĐLBT điện tich : $n_{NO_3^-} =t= 3z+0,1-0,02.2 = 0,1
2$ mol