Đáp án:
\(\begin{array}{l}
Magie(Mg),Ca(Can\,xi)\\
m = 19,6g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
RC{O_3} + {H_2}S{O_4} \to RS{O_4} + C{O_2} + {H_2}O\\
{n_{C{O_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{RC{O_3}}} = {n_{C{O_2}}} = 0,15\,mol\\
{M_{RC{O_3}}} = \dfrac{{14,2}}{{0,15}} = 94,67(g/mol)\\
\Rightarrow {M_R} = 94,67 - 60 = 34,67(g/mol)\\
\Rightarrow Magie(Mg),Ca(Can\,xi)\\
hh:MgC{O_3}(a\,mol),CaC{O_3}(b\,mol)\\
84a + 100b = 14,2\\
a + b = 0,15\\
\Rightarrow a = 0,05;b = 0,1\,\\
{n_{MgS{O_4}}} = {n_{MgC{O_3}}} = 0,05\,mol\\
{n_{CaS{O_4}}} = {n_{CaC{O_3}}} = 0,1\,mol\\
m = 0,05 \times 120 + 0,1 \times 136 = 19,6g
\end{array}\)