Đáp án:
\(\begin{array}{l}
a)m = 18,4g\\
b){m_{CaC{l_2}}} = 27,75g\\
c)\\
\% {m_{CaO}} = 45,65\% \\
\% {m_{CaC{O_3}}} = 54,35\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
CaO + 2HCl \to CaC{l_2} + {H_2}O(1)\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O(2)\\
a)\\
{n_{C{O_2}}} = 0,1mol\\
{n_{HCl}} = 0,5mol\\
\to {n_{HCl(2)}} = 2{n_{C{O_2}}} = 0,2mol\\
\to {n_{HCl(1)}} = 0,3mol\\
\to {n_{CaO}} = \dfrac{1}{2}{n_{HCl(1)}} = 0,15mol\\
\to {m_{CaO}} = 8,4g\\
\to {n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,1mol\\
\to {m_{CaC{O_3}}} = 10g\\
\to m = 8,4 + 10 = 18,4g\\
b)\\
{n_{CaC{l_2}}} = {n_{CaO}} + {n_{CaC{O_3}}} = 0,25mol\\
\to {m_{CaC{l_2}}} = 27,75g\\
c)\\
\% {m_{CaO}} = \dfrac{{8,4}}{{18,4}} \times 100\% = 45,65\% \\
\% {m_{CaC{O_3}}} = 100\% - 45,65\% = 54,35\%
\end{array}\)
