Giải thích các bước giải:
\(\begin{array}{l}
3.\\
a)CaC{l_2} + 2AgN{O_3} \to 2AgCl + Ca{(N{O_3})_2}\\
b)\\
{n_{CaC{l_2}}} = 0,3mol\\
{n_{AgN{O_3}}} = 0,2mol\\
\to {n_{AgN{O_3}}} < {n_{CaC{l_2}}} \to {n_{CaC{l_2}}}dư\\
\to {n_{AgCl}} = {n_{AgN{O_3}}} = 0,2mol\\
\to {m_{AgCl}} = 28,7g\\
c)\\
{n_{Ca{{(N{O_3})}_2}}} = \dfrac{1}{2}{n_{AgN{O_3}}} = 0,1mol\\
{n_{CaC{l_2}(pt)}} = \dfrac{1}{2}{n_{AgN{O_3}}} = 0,1mol\\
\to {n_{CaC{l_2}(dư)}} = 0,2mol\\
\to C{M_{CaC{l_2}(dư)}} = \dfrac{{0,2}}{{0,05 + 0,15}} = 1M\\
\to C{M_{Ca{{(N{O_3})}_2}}} = \dfrac{{0,1}}{{0,05 + 0,15}} = 0,5M
\end{array}\)
\(\begin{array}{l}
4.\\
BaC{l_2} + {H_2}S{O_4} \to BaS{O_4} + 2HCl\\
{n_{{H_2}S{O_4}}} = 0,05mol\\
{n_{BaC{l_2}}} = 0,03mol\\
\to {n_{BaC{l_2}}} < {n_{{H_2}S{O_4}}} \to {n_{{H_2}S{O_4}}}dư\\
a)\\
{n_{BaS{O_4}}} = {n_{BaC{l_2}}} = 0,03mol\\
\to {m_{BaS{O_4}}} = 6,99g\\
b)\\
{n_{HCl}} = 2{n_{BaC{l_2}}} = 0,06mol\\
{n_{{H_2}S{O_4}(pt)}} = {n_{BaC{l_2}}} = 0,03mol\\
\to {n_{{H_2}S{O_4}(dư)}} = 0,02mol\\
\to C{M_{{H_2}S{O_4}(dư)}} = \dfrac{{0,02}}{{0,03 + 0,02}} = 0,4M\\
\to C{M_{HCl}} = \dfrac{{0,06}}{{0,03 + 0,02}} = 1,2M
\end{array}\)
