Giải thích các bước giải:

\[\begin{array}{l}
{n_{{A_2}S{O_4}}} = a\;{\rm{mol}};{{\rm{n}}_{{\rm{BS}}{{\rm{O}}_{\rm{4}}}}}{\rm{ = b}}\;{\rm{mol}}\\
{A_2}S{O_4} + BaC{l_2} \to BaS{O_4} + 2ACl\\
BS{O_4} + BaC{l_2} \to BaS{O_4} + BC{l_2}\\
{n_{BaS{O_4}}} = \frac{{6,99}}{{233}} = 0,03\;{\rm{mol}}\\
BTNT\;{\rm{S}}:\\
{n_{BaC{l_2}}} = {n_{BaS{O_4}}} = 0,03\;{\rm{mol}}\\
{\rm{BTKL:}}\\
 \Rightarrow {m_Y}{\rm{ }} = {\rm{ }}3,82{\rm{ }} + {\rm{ }}0,03.208{\rm{ }} - {\rm{ }}6,99{\rm{ }} = {\rm{ }}3,07{\mkern 1mu} {\kern 1pt} {\rm{gam}}\\
\begin{array}{*{20}{l}}
{{n_X} = {n_{BaS{O_4}}} = 0,03mol}\\
{ \Rightarrow {M_X} = \frac{{3,82}}{{0,03}} = 127}
\end{array}\\
\begin{array}{*{20}{l}}
{B = A + 1}\\
{2A + 96 > 127 > B + 96}
\end{array}\\
 \Rightarrow 15,5 < A < 30\\
 \Rightarrow A = 23\;(Na);B = 24\;(Mg)
\end{array}\]