- Từ giả thiết tính được : \({n_{C{l_2}}} = 0,035mol; {n_{{O_2}}} = 0,025mol\)
Theo ĐLBT khối lượng :
\(\begin{array}{l}
{m_{C{l_2}}} + {m_{{O_2}}} + {m_{Al}} + {m_{Mg}} = 5,055 \Rightarrow {m_{Al}} + {m_{Mg}} = 5,055 - 3,285 = 1,77gam\\
\left. \begin{array}{l}
Mg \to M{g^{2 + }} + 2{\rm{e}}\\
x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2{\rm{x}}\\
Al \to A{l^{3 + }} + 3{\rm{e}}\\
y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3y
\end{array} \right\};\\
\left. \begin{array}{l}
C{l_2}\,\,\,\,\,\,\,\, + \,\,\,\,\,2{\rm{e}} \to 2C{l^ - }\\
0,035\,\,\,\,\,\,0,07\,\,\,\,\,\left( {mol} \right)\\
{O_2}\,\,\,\,\,\,\,\, + \,\,\,\,\,\,4e \to 2{O^{2 - }}\\
0.025\,\,\,\,\,\,\,\,\,0,1\,\,\,\,\,\,\,\left( {mol} \right)
\end{array} \right\}2{\rm{x}} + 3y = 0,17\,\,(3)
\end{array}\)
Từ (3) và (4)
\( \Rightarrow \left\{ \begin{array}{l}
x = 0,04\\
y = 0,03
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{m_{Mg}} = 0,04 \times 24 = 0,96\,(gam)\\
{m_{Al}} = 0,03 \times 27 = 0,81\,(gam)
\end{array} \right.\)
