Đáp án + Giải thích các bước giải:

$a\big)$

Bảo toàn e: $n_{Cu}=n_{SO_2}=\frac{8,96}{22,4}=0,4(mol)$

$\to \%m_{Cu}=\frac{0,4.64}{41,6}.100\%\approx 61,54\%$

$\to \%m_{CuO}=100-61,54=38,46\%$

$b\big)$

$n_{CuO}=\frac{41,6-0,4.64}{80}=0,2(mol)$

Bảo toàn e: $n_{H_2SO_4}=2n_{Cu}+2n_{CuO}=1,2(mol)$

$\to C\%_{H_2SO_4}=\frac{1,2.98}{120.1,98}.100\%\approx 49,49\%$

Em tham khảo nha :

\(\begin{array}{l}
a)\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
{n_{S{O_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4mol\\
{n_{Cu}} = {n_{S{O_2}}} = 0,4mol\\
{m_{Cu}} = 0,4 \times 64 = 25,6g\\
{m_{CuO}} = 41,6 - 25,6 = 16g\\
\% Cu = \dfrac{{25,6}}{{41,6}} \times 100\%  = 61,54\% \\
\% CuO = 100 - 61,54 = 38,46\% \\
b)\\
{n_{CuO}} = \dfrac{{16}}{{80}} = 0,2mol\\
{n_{{H_2}S{O_4}}} = 2{n_{Cu}} + {n_{CuO}} = 1mol\\
{m_{{H_2}S{O_4}}} = 1 \times 98 = 98g\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = 120 \times 1,98 = 237,6g\\
C{\% _{{H_2}S{O_4}}} = \dfrac{{98}}{{237,6}} \times 100\%  = 41,25\% 
\end{array}\)