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: 

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