Đáp án:
Bạn tham khảo lời giải ở dưới nhé!!!
Giải thích các bước giải:
\(\begin{array}{l}
CaO + 2HCl \to CaC{l_2} + {H_2}O\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
{n_{C{O_2}}} = 0,2mol\\
\to {n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,2mol\\
\to {m_{CaC{O_3}}} = 20g\\
\to {m_{CaO}} = 5,6g\\
\to {n_{CaO}} = 0,1mol\\
\to \% {m_{CaC{O_3}}} = \dfrac{{20}}{{25,6}} \times 100\% = 78,125\% \\
\to \% {m_{CaO}} = 21,875\% \\
{n_{HCl}} = 2({n_{CaO}} + {n_{CaC{O_3}}}) = 0,6mol\\
\to {m_{HCl}} = 21,9g\\
\to {m_{HCl}}{\rm{dd = }}\dfrac{{21,9 \times 100}}{{14,6}} = 150g\\
{n_{CaC{l_2}}} = {n_{CaO}} + {n_{CaC{O_3}}} = 0,3mol\\
\to {m_{CaC{l_2}}} = 33,3g\\
{m_{{\rm{dd}}}} = {m_A} + {m_{HCl}}{\rm{dd}} - {m_{C{O_2}}} = 25,6 + 150 - 0,2 \times 44 = 166,8g\\
\to C{\% _{CaC{l_2}}} = \dfrac{{33,3}}{{166,8}} \times 100\% = 19,96\%
\end{array}\)
