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Đáp án:

\(\begin{array}{l}
a)\\
a = 16,1g\\
b)\\
m = 26,2g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Ba + 2{H_2}O \to Ba{(OH)_2} + {H_2}(1)\\
Ba + 2HCl \to BaC{l_2} + {H_2}(2)\\
Mg + 2HCl \to MgC{l_2} + {H_2}(3)\\
{n_{{H_2}(1)}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol \Rightarrow {n_{Ba}} = {n_{{H_2}(1)}} = 0,1\,mol\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
{n_{Mg}} = 0,2 - 0,1 = 0,1\,mol\\
a = 0,1 \times 137 + 0,1 \times 24 = 16,1g\\
b)\\
BaC{l_2} + {H_2}S{O_4} \to BaS{O_4} + 2HCl(4)\\
NaOH + HCl \to NaCl + {H_2}O(5)\\
MgC{l_2} + 2NaOH \to Mg{(OH)_2} + 2NaCl(6)\\
{n_{BaS{O_4}}} = {n_{BaC{l_2}}} = 0,1\,mol\\
{n_{HCl}} = 2{n_{BaC{l_2}}} = 0,2\,mol\\
{n_{NaOH(5)}} = {n_{HCl}} = 0,2\,mol\\
{n_{NaOH(6)}} = 0,3 - 0,2 = 0,1\,mol\\
{n_{Mg{{(OH)}_2}}} = \dfrac{{0,1}}{2} = 0,05\,mol\\
m = 0,1 \times 233 + 0,05 \times 58 = 26,2g
\end{array}\)