Đáp án:
\(\begin{array}{l}
{m_{CuO}} = 4,8g\\
{m_{F{e_x}{O_y}}} = 9,6g\\
CTHH:F{e_2}{O_3}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
CuO + {H_2} \xrightarrow{t^0} Cu + {H_2}O\\
F{e_x}{O_y} + y{H_2} \xrightarrow{t^0} xFe + y{H_2}O\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{2,688}}{{22,4}} = 0,12\,mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,12\,mol\\
{m_{Cu}} = 10,56 - 0,12 \times 56 = 3,84g\\
{n_{Cu}} = \dfrac{{3,84}}{{64}} = 0,06\,mol\\
{n_{CuO}} = {n_{Cu}} = 0,06\,mol\\
{m_{F{e_x}{O_y}}} = 14,4 - 0,06 \times 80 = 9,6g\\
{m_O} = 9,6 - 0,12 \times 56 = 2,88g\\
{n_O} = \dfrac{{2,88}}{{16}} = 0,18\,mol\\
x:y = {n_{Fe}}:{n_O} = 0,12:0,18 = 2:3\\
\Rightarrow CTHH:F{e_2}{O_3}
\end{array}\)
\({m_{CuO}} = 0,06 \times 80 = 4,8g\)