Đáp án:
2,24l
Giải thích các bước giải:
\(\begin{array}{l}
hh:Cu(a\,mol),CuO(b\,mol)\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b\\
nCuS{O_4} = \dfrac{{32}}{{160}} = 0,2\,mol\\
64a + 80b = 14,4\\
a + b = 0,2\\
\Rightarrow a = b = 0,1\\
nS{O_2} = nCu = 0,1\,mol\\
VS{O_2} = 0,1 \times 22,4 = 2,24l\,
\end{array}\)