Giải thích các bước giải:
\(\begin{array}{l}
1.\\
NaOH + HCl \to NaCl + {H_2}O(1)\\
Ca{(OH)_2} + 2HCl \to CaC{l_2} + 2{H_2}O(2)\\
{n_{HCl}} = 0,15mol
\end{array}\)
Gọi a và b lần lượt là số mol của HCl(1) và HCl(2)
\(\begin{array}{l}
\left\{ \begin{array}{l}
a + b = 0,15\\
58,5a + 55,5b = 8,715
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,13\\
b = 0,02
\end{array} \right.\\
\to {n_{HCl}}(1) = 0,13mol\\
\to {n_{HCl}}(2) = 0,02mol\\
a)\\
{n_{NaOH}} = {n_{HCl}}(1) = 0,13mol\\
{n_{Ca{{(OH)}_2}}} = \dfrac{1}{2}{n_{HCl}}(2) = 0,01mol\\
\to C{M_{NaOH}} = \dfrac{{0,13}}{{0,1}} = 1,3M\\
\to C{M_{Ca{{(OH)}_2}}} = \dfrac{{0,01}}{{0,1}} = 0,1M\\
b)\\
{n_{NaCl}} = {n_{HCl}}(1) = 0,13mol \to {m_{NaCl}} = 7,605g\\
{n_{CaC{l_2}}} = \dfrac{1}{2}{n_{HCl}}(2) = 0,01mol \to {m_{CaC{l_2}}} = 1,11g\\
\to \% {m_{NaCl}} = \dfrac{{7,605}}{{8,715}} \times 100\% = 87,26\% \\
\to \% {m_{CaC{l_2}}} = \dfrac{{1,11}}{{8,715}} \times 100\% = 12,74\% \\
2.\\
CaO + {H_2}O \to Ca{(OH)_2}\\
{n_{CaO}} = 0,05mol\\
a)\\
{n_{Ca{{(OH)}_2}}} = {n_{CaO}} = 0,05mol\\
\to C{M_{Ca{{(OH)}_2}}} = \dfrac{{0,05}}{4} = 0,0125M\\
b)\\
{n_{S{O_2}}} = 0,1mol\\
\to \dfrac{{{n_{Ca{{(OH)}_2}}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,05}}{{0,1}} = 0,5
\end{array}\)
=> Tạo 1 muối: CaSO3
\(\begin{array}{l}
S{O_2} + Ca{(OH)_2} \to CaS{O_3} + {H_2}O\\
{n_{CaS{O_3}}} = {n_{S{O_2}}} = 0,1mol\\
\to {m_{CaS{O_3}}} = 12g
\end{array}\)
