Đáp án:
\(\begin{array}{l}
C{\% _{CaC{l_2}}} = 10,51\% \\
C{\% _{HCl}} = 1,73\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
b)\\
{n_{CaC{O_3}}} = \dfrac{m}{M} = \dfrac{{20}}{{100}} = 0,2mol\\
{m_{HCl}} = \dfrac{m}{M} = \dfrac{{200 \times 9,125\% }}{{100\% }} = 18,25g\\
{n_{HCl}} = \dfrac{{18,25}}{{36,5}} = 0,5mol\\
\dfrac{{0,2}}{1} < \dfrac{{0,5}}{2} \Rightarrow HCl\text{ dư}\\
{n_{C{O_2}}} = {n_{CaC{O_3}}} = 0,2mol\\
{m_{C{O_2}}} = n \times M = 0,2 \times 44 = 8,8g\\
{m_{ddspu}} = {m_{CaC{O_3}}} + {m_{ddHCl}} - {m_{C{O_2}}} = 20 + 200 - 8,8 = 211,2g\\
{n_{CaC{l_2}}} = {n_{CaC{O_3}}} = 0,2mol\\
{m_{CaC{l_2}}} = n \times M = 0,2 \times 111 = 22,2g\\
{n_{HCld}} = {n_{HCl}} - 2{n_{CaC{O_3}}} = 0,5 - 2 \times 0,2 = 0,1mol\\
{m_{HCl}} = n \times M = 0,1 \times 36,5 = 3,65g\\
C{\% _{CaC{l_2}}} = \dfrac{{{m_{CaC{l_2}}}}}{{{m_{{\rm{dd}}spu}}}} \times 100\% = \dfrac{{22,2}}{{211,2}} \times 100\% = 10,51\% \\
C{\% _{HCl}} = \dfrac{{{m_{HCl}}}}{{{m_{{\rm{dd}}spu}}}} \times 100\% = \dfrac{{3,65}}{{211,2}} \times 100\% = 1,73\%
\end{array}\)
