Giải thích các bước giải:
\(\begin{array}{l}
a)BaC{l_2} + N{a_2}S{O_4} \to BaS{O_4} + 2NaCl\\
{m_{BaC{l_2}}} = \dfrac{{200 \times 10,8\% }}{{100\% }} = 21,6g\\
\to {n_{BaC{l_2}}} = 0,104mol\\
{m_{N{a_2}S{O_4}}} = \dfrac{{100 \times 14,2\% }}{{100\% }} = 14,2g\\
\to {n_{N{a_2}S{O_4}}} = 0,1mol\\
\to {n_{BaC{l_2}}} > {n_{N{a_2}S{O_4}}} \to {n_{BaC{l_2}}}dư\\
\to {n_{BaC{l_2}}}(pt) = {n_{N{a_2}S{O_4}}} = 0,1mol\\
\to {n_{BaC{l_2}}}(dư) = 0,004mol \to {m_{BaC{l_2}}}(dư) = 0,376g\\
b)\\
{n_{BaS{O_4}}} = {n_{BaC{l_2}}} = 0,104mol\\
\to {m_{BaS{O_4}}} = 24,232g\\
c)\\
{n_{NaCl}} = 2{n_{N{a_2}S{O_4}}} = 0,2mol\\
\to {m_{NaCl}} = 11,7g\\
\to {m_{{\rm{dd}}}} = {m_{{\rm{dd}}BaC{l_2}}} + {m_{{\rm{dd}}N{a_2}S{O_4}}} - {m_{BaS{O_4}}} = 275,8g\\
\to C{\% _{NaCl}} = \dfrac{{11,7}}{{275,8}} \times 100\% = 4,24\% \\
\to C{\% _{BaC{l_2}}} = \dfrac{{0,376}}{{275,8}} \times 100\% = 0,14\%
\end{array}\)
