Em tham khảo nha :
\(\begin{array}{l}
a)\\
BaO + {H_2}S{O_4} \to BaS{O_4} + {H_2}O\\
{n_{BaO}} = \dfrac{{15,3}}{{153}} = 0,1mol\\
{n_{{H_2}S{O_4}}} = {n_{BaO}} = 0,1mol\\
{m_{{H_2}S{O_4}}} = 0,1 \times 98 = 9,8g\\
C{\% _{{H_2}S{O_4}}} = \dfrac{{9,8}}{{100}} \times 100\% = 9,8\% \\
b)\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{NaOH}} = 2{n_{{H_2}S{O_4}}} = 0,2mol\\
{m_{NaOH}} = 0,2 \times 40 = 8g\\
c)\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
{n_{Fe}} = \dfrac{{11,2}}{{56}} = 0,2mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{10 \times 98\% }}{{98}} = 0,1mol\\
\dfrac{{0,2}}{2} > \dfrac{{0,1}}{6} \Rightarrow Fe\text{ dư}\\
{n_{S{O_2}}} = \dfrac{{{n_{{H_2}S{O_4}}}}}{2} = 0,05mol\\
{V_{S{O_2}}} = 0,05 \times 22,4 = 1,12l
\end{array}\)