\(n_{Zn}=\dfrac{2,6}{65}=0,04mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,04 0,08 0,04 0,04 ( mol )
\(m_{ZnCl_2}=0,04.136=5,44g\)
\(V_{H_2}=0,04.22,4=0,896l\)
\(C_{M_{HCl}}=\dfrac{0,08}{0,1}=0,8M\)
\(m_{HCl}=0,08.36,5=2,92g\)
\(m_{ddHCl}=100.1,198=119,8g\)
\(C\%_{HCl}=\dfrac{2,92}{119,8}.100=2,43\%\)
