\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)
\(\Rightarrow m_{H_2}=0,35.2=0,7\left(g\right)\)
Theo pthh ta suy ra được: \(n_{HCl}=2n_{H_2}\)
\(\Rightarrow n_{HCl}=0,35.2=0,7\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,7.36,5=25,55\left(g\right)\)
Áp dụng đlBTKL có:
\(m_{hh}+m_{HCl}=m_{muối}+m_{H_2}\)
\(m_{hh}+25,55=39,65+0,7\)
\(\Rightarrow m=\left(39,65+0,7\right)-25,55=14,77\left(g\right)\)
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