MgO dư nhé :>

Em tham khảo nha:

\(\begin{array}{l}
1)\\
a)\\
ZnO + 2HCl \to ZnC{l_2} + {H_2}O\\
{m_{HCl}} = \dfrac{{{m_{{\rm{dd}}}} \times {C_\% }}}{{100}} = \dfrac{{120 \times 6}}{{100}} = 7,2g\\
{n_{HCl}} = \dfrac{m}{M} = \dfrac{{7,2}}{{36,5}} \approx 0,2\,mol\\
{n_{ZnO}} = \dfrac{{0,2}}{2} = 0,1\,mol\\
{m_{ZnO}} = n \times M = 0,1 \times 81 = 8,1g\\
b)\\
{m_{{\rm{dd}}spu}} = 8,1 + 120 = 128,1g\\
{n_{ZnC{l_2}}} = {n_{ZnO}} = 0,1\,mol\\
{C_\% }ZnC{l_2} = \dfrac{m}{{{m_{{\rm{dd}}}}}} \times 100\%  = \dfrac{{0,1 \times 136}}{{128,1}} \times 100\%  = 10,62\% \\
2)\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
{n_{MgO}} = \dfrac{m}{M} = \frac{{12}}{{40}} = 0,3\,mol\\
{m_{HCl}} = \dfrac{{{m_{{\rm{dd}}}} \times {C_\% }}}{{100}} = \dfrac{{146 \times 10}}{{100}} = 14,6g\\
{n_{HCl}} = \dfrac{m}{M} = \frac{{14,6}}{{36,5}} = 0,4\,mol\\
\dfrac{{0,3}}{1} > \dfrac{{0,4}}{2} \Rightarrow \text{ MgO dư}\\
{n_{MgO}} \text{ phản ứng}= \dfrac{{0,4}}{2} = 0,2\,mol\\
{n_{MgC{l_2}}} = {n_{MgO}} = 0,2\,mol\\
{m_{{\rm{dd}}}} = 146 + 0,2 \times 40 = 154g\\
{C_\% }MgC{l_2} = \dfrac{{0,2 \times 95}}{{154}} \times 100\%  = 12,34\% 
\end{array}\)