Đáp án:
\(\begin{array}{l}
a)\\
{m_{Al}} = 2,7g\\
{m_{Mg}} = 3,6g\\
b)\\
CTHH:CuO
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
hh:Al(a\,mol),Mg(b\,mol)\\
\left\{ \begin{array}{l}
27a + 24b = 6,3\\
\frac{3}{2}a + b = 0,3
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,15\\
{m_{Al}} = 0,1 \times 27 = 2,7g\\
{m_{Mg}} = 6,3 - 2,7 = 3,6g\\
b)\\
{M_x}{O_y} + y{H_2} \to xM + y{H_2}O\\
{n_{{M_x}{O_y}}} = \dfrac{{{n_{{H_2}}}}}{y} = \dfrac{{0,3}}{y}\\
{M_{{M_x}{O_y}}} = 24,1:\dfrac{{0,3}}{y} = \dfrac{{241}}{3}y\\
x{M_M} + 16y = \dfrac{{241}}{3}y\\
{M_M} = \dfrac{{(80,33 - 16)y}}{x}\\
y = 1 \Rightarrow x = 1\\
\Rightarrow {M_M} = 64dvC\\
\Rightarrow M:\text{Đồng}(Cu)\\
CTHH:CuO
\end{array}\)
