Giải thích các bước giải:
\(\begin{array}{l}
CaS + 2HCl \to CaC{l_2} + {H_2}S\\
a)\\
{n_{{H_2}S}} = 0,03mol\\
\to {n_{HCl}} = 2{n_{{H_2}S}} = 0,06mol \to {m_{HCl}} = 2,19g\\
\to {m_1} = \dfrac{{2,19}}{{8,58\% }} \times 100\% = 25,5\% \\
{n_{CaC{l_2}}} = {n_{{H_2}S}} = 0,03mol \to {m_{CaC{l_2}}} = 3,33g\\
\to {m_2} = \dfrac{{3,33}}{{9,6\% }} \times 100\% = 34,69\% \\
{n_{{\rm{CaS}}}} = {n_{{H_2}S}} = 0,03mol \to {m_{{\rm{CaS}}}} = 2,18g
\end{array}\)
