\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{HCl}=0,2.2=0,4\left(mol\right)\\ m_{ddHCl}=\dfrac{0,4.36,5.100}{10}=146\left(g\right)\\ n_{H_2}=n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ C\%_{ddZnCl_2}=\dfrac{136.0,2}{13+146-0,2.2}.100\approx17,15\%\)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: Zn + 2HCl → ZnCl2 + H2
Mol: 0,2 0,4 0,2 0,2
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
\(m_{ddHCl}=\dfrac{0,4.36,5.100}{10}=146\left(g\right)\)
mdd sau pứ = 13+146-0,2.2 = 158,6 (g)
\(C\%_{ddZnCl_2}=\dfrac{0,2.136.100\%}{158,6}=17,15\%\)
