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$2K+2H_2O\rightarrow 2KOH+H_2\uparrow$
$x 0,5x$
$2KOH+2Al+2H_2O\rightarrow KAlO_2+ 3H_2\uparrow $
$x 1,5x$
$2K+2HCl\rightarrow 2KCl+H_2\uparrow$
$x 0,5x$
$2Al+6HCl\rightarrow 2AlCl_3+ 3H_2\uparrow $
$y 1,5y$
$\begin{cases}2x=\frac{4,48}{22,4} \\ 0,5x+1,5y=\frac{6,72}{22,4} \end{cases}\Rightarrow \begin{cases}x=0,1 mol \\ y=\frac{0,5}{3} mol \end{cases}\Rightarrow m=39.0,1+27.
\frac{0,5}{3} =8,4 gam$
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Trả lời 19-01-13 05:23 PM
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