a) PTHH:

MgO + 2HCl → MgCl₂ + H₂O

MgCO₃ + 2HCl → MgCl₂ + H₂O + CO₂↑

\[\begin{array}{l}{n_{C{O_2}}} = \frac{{2,24}}{{22,4}} = 0,1(mol)\\{n_{MgC{O_3}}} = {n_{C{O_2}}} = 0,1(mol)\\{m_{MgC{O_3}}} = 0,1.84 = 8,4(g)\\{m_{MgO}} = 16,4 - 8,4 = 8(g)\end{array}\]

b)

\[\begin{array}{l}{n_{MgO}} = \frac{8} = 0,2(mol)\\{n_{HCl}} = 2{n_{MgO}} + 2{n_{C{O_2}}} = 2.0,2 + 2.0,1 = 0,6(mol)\\{m_{HCl}} = 0,6.36,5 = 21,9(g)\\C{\% _{HCl}} = \frac{{21,9}}.100\% = 10,95\% \end{array}\]

c) \[{m_{{\rm{dd}}X}} = {m_{hh}} + {m_{{\rm{dd}}HCl}} - {m_{C{O_2}}} = 16,4 + 200 - 4,4 = 212(g)\]

\[\begin{array}{l}{n_{MgC{l_2}}} = {n_{MgO}} + {n_{C{O_2}}} = 0,1 + 0,1 = 0,2(mol)\\{m_{MgC{l_2}}} = 0,3.95 = 28,5(g)\\C{\% _{MgC{l_2}}} = \frac{{28,5}}.100\% = 13,44\% \end{array}\]