Đáp án:
Bạn tham khảo lời giải ở dưới nhé !
Giải thích các bước giải:
\(\begin{array}{l}
6.\\
2KOH + {H_2}S{O_4} \to {K_2}S{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = 0,15mol\\
a.\\
{n_{{K_2}S{O_4}}} = {n_{{H_2}S{O_4}}} = 0,15mol\\
\to {m_{{K_2}S{O_4}}} = 26,1g\\
b.\\
{n_{KOH}} = 2{n_{{H_2}S{O_4}}} = 0,3mol\\
\to {V_{KOH}} = \dfrac{{0,3}}{{0,2}} = 1,5l\\
c.\\
C{M_{{K_2}S{O_4}}} = \dfrac{{0,15}}{{0,3 + 1,5}} = 0,083M
\end{array}\)
\(\begin{array}{l}
8.\\
a.2AgN{O_3} + CuC{l_2} \to 2AgCl + Cu{(N{O_3})_2}\\
b.\\
{n_{Cu{{(N{O_3})}_2}}} = 0,2mol\\
\to {n_{AgCl}} = 2{n_{Cu{{(N{O_3})}_2}}} = 0,4mol\\
\to {m_{AgCl}} = 57,4g
\end{array}\)
\(\begin{array}{l}
c.\\
{n_{CuC{l_2}}} = {n_{Cu{{(N{O_3})}_2}}} = 0,2mol\\
\to {m_{CuC{l_2}}} = 27g\\
\to {m_{{\rm{dd}}CuC{l_2}}} = \dfrac{{27}}{{13,5\% }} \times 100\% = 200g
\end{array}\)
\(\begin{array}{l}
d.\\
{m_{{\rm{dd}}}} = {m_{{\rm{dd}}AgN{O_3}}} + {m_{{\rm{dd}}CuC{l_2}}} - {m_{AgCl}} = 392,6g\\
\to C{\% _{Cu{{(N{O_3})}_2}}} = \dfrac{{37,6}}{{392,6}} \times 100\% = 9,58\%
\end{array}\)
\(\begin{array}{l}
9.\\
CaC{l_2} + 2AgN{O_3} \to Ca{(N{O_3})_2} + 2AgCl\\
{n_{CaC{l_2}}} = 0,01mol\\
{n_{AgN{O_3}}} = 0,005mol\\
\to {n_{AgN{O_3}}} < {n_{CaC{l_2}}} \to {n_{CaC{l_2}}}dư
\end{array}\)
\(\begin{array}{l}
a.\\
{n_{AgCl}} = {n_{AgN{O_3}}} = 0,005mol\\
\to {m_{AgCl}} = 0,7175g\\
b.\\
{n_{Ca{{(N{O_3})}_2}}} = \dfrac{1}{2}{n_{AgN{O_3}}} = 0,0025mol\\
\to C{M_{Ca{{(N{O_3})}_2}}} = \dfrac{{0,0025}}{{0,25}} = 0,01M
\end{array}\)
