Em tham khảo nha :

\(\begin{array}{l}
a)\\
{n_{S{O_2}}} = \dfrac{{2,688}}{{22,4}} = 0,12mol\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
{n_{Cu}} = {n_{S{O_2}}} = 0,12mol\\
{m_{Cu}} = 0,12 \times 64 = 7,68g\\
{m_{CuO}} = 10,08 - 7,68 = 2,4g\\
b)\\
BaC{l_2} + {H_2}S{O_4} \to BaS{O_4} + 2HCl\\
BaC{l_2} + CuS{O_4} \to BaS{O_4} + CuC{l_2}\\
{n_{BaS{O_4}}} = \dfrac{{58,25}}{{233}} = 0,25mol\\
{n_{{H_2}S{O_4}d}} = 0,25 - 0,12 - 0,03 = 0,1mol\\
{n_{{H_2}S{O_4}}} = 0,25mol\\
{m_{{H_2}S{O_4}}} = 0,25 \times 98 = 24,5g\\
{m_{{\rm{dd}}{{\rm{H}}_2}S{O_4}}} = \dfrac{{24,5 \times 100}}{{98}} = 25g\\
c)\\
2S{O_2} + {O_2} \to 2S{O_3}\\
\dfrac{{0,12}}{2} < \dfrac{{0,08}}{1} \Rightarrow {O_2}\text{ dư}\\
{n_{{O_2}d}} = 0,08 - \frac{{0,12}}{2} = 0,02mol\\
{M_{hh}} = 33,68 \times 2 = 67,36dvC\\
 \Rightarrow \dfrac{{80{n_{S{O_3}td}} + 0,02 \times 32}}{{{n_{S{O_3}td}} + 0,02}} = 67,36\\
 \Rightarrow {n_{S{O_3}td}} = 0,06mol\\
{n_{S{O_3}}} = {n_{S{O_2}}} = 0,12mol\\
H = \dfrac{{0,06}}{{0,12}} \times 100\%  = 50\% 
\end{array}\)