Đáp án:

a) 28,89% và 71,11%

b) 2,17 ml

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
b)\\
hh:Zn(a\,mol),CuO(b\,mol)\\
{n_{{H_2}S{O_4}}} = 0,12 \times 0,5 = 0,06\,mol\\
\left\{ \begin{array}{l}
65a + 80b = 4,5\\
a + b = 0,06
\end{array} \right.\\
 \Rightarrow a = 0,02;b; =  0,04
\end{array}\)

\(\begin{array}{l}
\% {m_{Zn}} = \frac{{0,02 \times 65}}{{4,5}} \times 100\%  = 28,89\% \\
\% {m_{CuO}} = 100 - 28,89 = 71,11\% \\
c)\\
{m_{Zn}} = 2,25 \times 28,89\%  = 0,65g\\
{m_{CuO}} = 2,25 - 0,65 = 1,6g\\
{n_{Zn}} = \dfrac{{0,65}}{{65}} = 0,01\,mol\\
{n_{CuO}} = \dfrac{{1,6}}{{80}} = 0,02\,mol\\
Zn + 2{H_2}S{O_4} \to ZnS{O_4} + S{O_2} + 2{H_2}O\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
{n_{{H_2}S{O_4}}} = 0,01 \times 2 + 0,02 = 0,04\,mol\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{0,04 \times 98}}{{98\% }} = 4g\\
{V_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{4}{{1,84}} = 2,17ml
\end{array}\)