\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)
Zn + 2HCl -- > ZnCl2 + H2
mZnCl2 = 0,3 . 136 = 40,8(g)
VH2 = 0,3.22,4 = 6,72(l)
\(n_{Fe_2O_3}=\dfrac{19,2}{160}=0,12\left(mol\right)\)
\(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
LTL : \(\dfrac{0,3}{3}< \dfrac{0,12}{1}\)
=> H2 chưa khử hết Fe2O3
=> \(m_{Fe}=\dfrac{2}{3}.0,3.56=11,2\left(g\right)\)