Đáp án:
Câu 3:
\(\begin{array}{l}
a.\\
\% {m_{MgC{O_3}}} = 84,28\% \\
\% {m_{MgO}} = 15,72\% \\
b.\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = 205,8g
\end{array}\)
Giải thích các bước giải:
Câu 3:
\(\begin{array}{l}
a.\\
MgO + {H_2}S{O_4} \to MgS{O_4} + {H_2}O\\
MgC{O_3} + {H_2}S{O_4} \to MgS{O_4} + C{O_2} + {H_2}O\\
{n_{C{O_2}}} = 0,3mol\\
\to {n_{MgC{O_3}}} = {n_{{H_2}}} = 0,3mol\\
\to {m_{MgC{O_3}}} = 25,2g\\
\to {m_{MgO}} = 4,7g \to {n_{MgO}} = 0,1175mol\\
\to \% {m_{MgC{O_3}}} = \dfrac{{25,2}}{{29,9}} \times 100\% = 84,28\% \\
\to \% {m_{MgO}} = 100\% - 84,28\% = 15,72\% \\
b.\\
{n_{{H_2}S{O_4}}} = {n_{MgC{O_3}}} + {n_{MgO}} = 0,42mol\\
\to {m_{{H_2}S{O_4}}} = 41,16g\\
\to {m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{41,16}}{{20\% }} \times 100\% = 205,8g
\end{array}\)