Đáp án:

\(\begin{array}{l}
13.\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = 326,67g\\
{m_{BaS{O_4}}} = 116,5g\\
14.\\
\% {m_{Mg}} = 12\% \\
\% {m_{MgO}} = 88\% \\
{m_{{\rm{dd}}HCl}} = 135g
\end{array}\) 

\(\begin{array}{l}
15.\\
{m_{NaOH(dư)}} = 2g\\
{m_{N{a_2}C{O_3}}} = 5,3g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
13.\\
a)Ba{(OH)_2} + {H_2}S{O_4} \to B{\rm{aS}}{O_4} + 2{H_2}O\\
b)\\
{n_{Ba{{(OH)}_2}}} = 0,5mol\\
 \to {n_{{H_2}S{O_4}}} = {n_{Ba{{(OH)}_2}}} = 0,5mol\\
 \to {m_{{H_2}S{O_4}}} = 49g\\
 \to {m_{{\rm{dd}}{H_2}S{O_4}}} = \frac{{49}}{{15\% }} \times 100\%  = 326,67g\\
c)\\
{n_{BaS{O_4}}} = {n_{Ba{{(OH)}_2}}} = 0,5mol\\
 \to {m_{BaS{O_4}}} = 116,5g\\
14.\\
a/\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
b/\\
{n_{{H_2}}} = 0,05mol\\
 \to {n_{Mg}} = {n_{{H_2}}} = 0,05mol\\
 \to {m_{Mg}} = 1,2g\\
 \to {m_{MgO}} = 8,8g \to {n_{MgO}} = 0,22mol\\
 \to \% {m_{Mg}} = \dfrac{{1,2}}{{10}} \times 100\%  = 12\% \\
 \to \% {m_{MgO}} = 100\%  - 12\%  = 88\% \\
c/\\
{n_{HCl}} = 2{n_{Mg}} + 2{n_{MgO}} = 0,54mol\\
 \to {m_{HCl}} = 19,71g\\
 \to {m_{{\rm{dd}}HCl}} = \dfrac{{19,71}}{{14,6\% }} \times 100\%  = 135g
\end{array}\)

\(\begin{array}{l}
15.\\
{n_{C{O_2}}} = 0,05mol\\
{n_{NaOH}} = 0,15mol\\
 \to \dfrac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = \dfrac{{0,15}}{{0,05}} = 3
\end{array}\)

=> Tạo 1 muối: \(N{a_2}C{O_3}\)

\(\begin{array}{l}
a)C{O_2} + 2NaOH \to N{a_2}C{O_3} + {H_2}O\\
b)\\
{n_{NaOH}} > {n_{C{O_2}}} \to {n_{NaOH}}dư\\
{n_{NaOH(pt)}} = 2{n_{C{O_2}}} = 0,1mol\\
 \to {n_{NaOH(dư)}} = 0,15 - 0,1 = 0,05mol\\
 \to {m_{NaOH(dư)}} = 2g\\
c)\\
{n_{N{a_2}C{O_3}}} = {n_{C{O_2}}} = 0,05mol\\
 \to {m_{N{a_2}C{O_3}}} = 5,3g
\end{array}\)