Hóa 8

N

ngobaotuan

Y

yui_2000

$\begin{matrix}
2M & + & 2nHCl & \rightarrow & 2MCl_n & + & H_2\\
0,6/n & \leftarrow & 0,6 & \rightarrow & 0,6/n & & 0,3
\end{matrix}$
a) $m_{H_2O}=71,8 \ (g)$
$m_{HCl}= 21,9 \ (g) \rightarrow n_{HCl}=0,6 \ (mol)$
$\frac{m_{MCl_n}}{m_{MCl_n}+m_{H_2O}}=0,254771$
$\Rightarrow m_{MCl_n}\approx 26,7 \ (g)$
Theo ĐLBTKL:
$m_{M}=m_{MCl_n}+m_{H_2}-m_{HCl}$
$\Leftrightarrow m\approx 26,7+0,3.2-21,9\approx 5,4$
b) $M\approx\frac{5,4}{0,6/n}\approx9n$
n|1|2|3
M|9 (loại)|18 (loại)|27 (Al)
Vậy M là nhôm (Al)
 
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