Em tham khảo nha:

\(\begin{array}{l}
a)\\
Ba + {H_2}S{O_4} \to BaS{O_4} + {H_2}(1)\\
{n_{Ba}} = \dfrac{{27,4}}{{137}} = 0,2\,mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{100 \times 9,8\% }}{{98}} = 0,1\,mol\\
{n_{Ba}} > {n_{{H_2}S{O_4}}} \Rightarrow Ba \text{ dư } \\
{n_{Ba}} \text{ dư } = 0,2 - 0,1 = 0,1\,mol\\
{n_{{H_2}(1)}} = {n_{BaS{O_4}}} = 0,1\,mol\\
Ba + 2{H_2}O \to Ba{(OH)_2} + {H_2}(2)\\
{n_{{H_2}(2)}} = {n_{Ba{{(OH)}_2}}} = 0,1\,mol\\
{n_{{H_2}}} = 0,1 + 0,1 = 0,2\,mol\\
{V_{{H_2}}} = 0,2 \times 22,4 = 4,48l\\
b)\\
{m_{{\rm{dd}}spu}} = 27,4 + 100 - 0,1 \times 233 - 0,2 \times 2 = 103,7g\\
{C_\% }Ba{(OH)_2} = \dfrac{{0,1 \times 171}}{{103,7}} \times 100\%  = 16,49\% 
\end{array}\)