Đáp án:
\(\begin{array}{l}
a)\\
m = 29g\\
b)\\
C{\% _{CaC{l_2}}} = 19,96\%
\end{array}\)
Giải thích các bước giải:
a)
\(\begin{array}{l}
CaO + 2HCl \to CaC{l_2} + {H_2}O(1)\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O(2)\\
{n_{C{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,15mol\\
{m_{CaC{O_3}}} = n \times M = 0,15 \times 100 = 15g\\
{m_{HCl}} = \dfrac{{200 \times 14,6}}{{100}} = 29,2g\\
{n_{HCl}} = \dfrac{m}{M} = \dfrac{{29,2}}{{36,5}} = 0,8mol\\
{n_{HCl(2)}} = 2{n_{C{O_2}}} = 0,3mol\\
\Rightarrow {n_{HCl(1)}} = {n_{HCl}} - {n_{HCl(2)}} = 0,8 - 0,3 = 0,5mol\\
{n_{CaO}} = \dfrac{{{n_{HCl(2)}}}}{2} = \dfrac{{0,5}}{2} = 0,25mol\\
{m_{CaO}} = n \times M = 0,25 \times 56 = 14g\\
m = {m_{CaC{O_3}}} + {m_{CaO}} = 15 + 14 = 29g\\
b)\\
{n_{CaC{l_2}}} = {n_{CaO}} + {n_{CaC{O_3}}} = 0,25 + 0,15 = 0,4mol\\
{m_{CaC{l_2}}} = n \times M = 0,4 \times 111 = 44,4g\\
{m_{ddspu}} = m + {m_{ddHCl}} - {m_{C{O_2}}} = 29 + 200 - 0,15 \times 44 = 222,4g\\
C{\% _{CaC{l_2}}} = \dfrac{{44,4}}{{222,4}} \times 100\% = 19,96\%
\end{array}\)
