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Các phương trình :
$Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow$
$M+2HCl\rightarrow MCl_2+H_2\uparrow$
a)Ta có : $n_{H_2}=\frac{0,672}{22,4}=0,03 mol\Rightarrow n_{HCl}=0,06 mol$
$\Rightarrow \overline{M}_{Zn,M}=\frac{1,7}{0,03}=56,6667\Rightarrow M<56,667$
$\Rightarrow m_{HCl 10\%}=0,06.36,5.\frac{100}{10}=21,9 gam$
Ta lại có:
$ \frac{1,7}{M}<\frac{1}{2}.\frac{36,5.10}{36,5.100}=0,05 mol\Rightarrow M>\frac{1,7}{0,05}\Rightarrow M>34$
$\Rightarrow 34<M<56,667\Rightarrow M=40\Rightarrow M:Ca$
b)Ta có :$\begin{cases}n_{Zn}+n_{Ca}=0,03 \\ 65.n_{Zn}+40.n_{Ca}=1,7 \end{cases}\Rightarrow \begin{cases}n_{Zn}=0,02 mol \\ n_{Ca}=0,01 mol \end{cases}$
$m_{dd B}=m_{HCl 10\%}+m_{Zn+Ca}-m_{H_2}=21,9+1,7-2.0,03=23,54 gam$
$\Rightarrow \begin{cases}C\%_{ZnCl_2}=\frac{0,02.136.100}{23,54}=11,555\% \\ C\%_{CaCl_2}=\frac{0,01.111.100}{23,54}=4,715\% \end{cases}$
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Trả lời 09-11-12 11:53 PM
viet130480
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