1) \(n_{HCl}=\dfrac{80.14,6\%}{36,5}=0,32\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,16<-0,32--->0,16--->0,16
a = 0,16.65 = 10,4 (g)
2) V = 0,16.22,4 = 3,584 (l)
3) mdd sau pư = 10,4 + 80 - 0,16.2 = 90,08 (g)
\(C\%_{ZnCl_2}=\dfrac{0,16.136}{90,08}.100\%=24,156\%\)