Đáp án:
\(\begin{array}{l}
{C_M}MgC{l_2} = 0,2M\\
{C_M}BaC{l_2} = 0,1M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
BaC{l_2} + N{a_2}S{O_4} \to BaS{O_4} + 2NaCl\\
{n_{N{a_2}S{O_4}}} = 0,12 \times 0,5 = 0,06\,mol\\
{n_{BaS{O_4}}} = \dfrac{{11,65}}{{233}} = 0,05\,mol\\
{n_{N{a_2}S{O_4}}} \text{ dư } = 0,06 - 0,05 = 0,01\,mol\\
{n_{BaC{l_2}}} = {n_{BaS{O_4}}} = 0,05\,mol\\
{n_{NaCl}} = 2{n_{BaS{O_4}}} = 0,1\,mol\\
\Rightarrow {m_{NaCl}} + {m_{N{a_2}S{O_4}}} \text{ dư }+ {m_{MgC{l_2}}} = 16,77g\\
\Rightarrow {m_{MgC{l_2}}} = 16,77 - 0,01 \times 142 - 0,1 \times 58,5 = 9,5g\\
{n_{MgC{l_2}}} = \dfrac{{9,5}}{{95}} = 0,1\,mol\\
{C_M}MgC{l_2} = \dfrac{{0,1}}{{0,5}} = 0,2M\\
{C_M}BaC{l_2} = \dfrac{{0,05}}{{0,5}} = 0,1M
\end{array}\)