|
|
$m_X:\begin{cases}n_{Na}=x mol \\n_{Al}=y mol \end{cases}$
$*)M+H_2O$:
$2Na+2H_2O\rightarrow 2NaOH+H_2\uparrow$
$x x 0,5x$
$2NaOH+2Al+2H_2O\rightarrow 2NaAlO_2+3H_2\uparrow$
$x x 1,5x$
$\Rightarrow n_{1 H_2}=0,5x+1,5x (1)$
$*)M+NaOH$:
$2Na+2H_2O\rightarrow 2NaOH+H_2\uparrow$
$x x 0,5x$
$2NaOH+2Al+2H_2O\rightarrow 2NaAlO_2+3H_2\uparrow$
$y y 1,5y$
$\Rightarrow n_{1 H_2}=0,5x+1,5y (2)$
$(1);(2)\Rightarrow \frac{n_{1 H_2}}{n_{2 H_2}}=\frac{0,5x+1,5x}{0,5x+1,5y}=\frac{5}{8,75}\Rightarrow y=2x$
$\Rightarrow \%m_{Na}=\frac{23x.100}{23x+27y}=
\frac{23x.100}{23x+27.2x}=29,87\% $
|
|
|
Trả lời 23-02-13 08:06 PM
|
|