Đáp án:
\(\begin{array}{l}
b/\\
\% {m_{CaC{O_3}}} = 64,1\% \\
\% {m_{CaO}} = 35,9\% \\
c/{V_{{\rm{dd}}HCl}} = 0,03l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a/\\
CaO + 2HCl \to CaC{l_2} + {H_2}O\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O
\end{array}\)
\(\begin{array}{l}
b/\\
{n_{C{O_2}}} = 0,05mol\\
\to {n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,05mol\\
\to {m_{CaC{O_3}}} = 5g\\
\to {m_{CaO}} = 7,8 - 5 = 2,8g \to {n_{CaO}} = 0,05mol\\
\to \% {m_{CaC{O_3}}} = \dfrac{5}{{7,8}} \times 100\% = 64,1\% \\
\to \% {m_{CaO}} = 100\% - 64,1\% = 35,9\%
\end{array}\)
\(\begin{array}{l}
c/\\
{n_{HCl}} = 2{n_{CaO}} + 2{n_{CaC{O_3}}} = 0,2mol\\
\to {m_{HCl}} = 7,3g\\
\to {m_{{\rm{dd}}HCl}} = \dfrac{{7,3}}{{20\% }} \times 100\% = 36,5g\\
\to {V_{{\rm{dd}}HCl}} = \dfrac{{36,5}}{{1,05}} = 34,76ml = 0,03l
\end{array}\)