$a.PTHH:Zn+2HCl→ZnCl_2+H_2$

$n_{Zn}=19,5:65=0,3(mol)$

$n_{H2}=n_{Zn}=0,3(mol)$

⇒$V_{H_2}=0,3.22,4=6,72(l)$

$b.PTHH:3H_2+Fe_2O_3\xrightarrow{t^o}2Fe+3H_2O$

$n_{Fe_2O_3}=19,2:160=0,12(mol)$

Lập tỉ lệ: $0,3:3<0,12:1→Fe_2O_3$ dư

$n_{Fe}=2:3n_{H_2}=0,2(mol)$

$m_{Fe}=0,2.56=11,2(g)$

$a,PTPƯ:Zn+2HCl\xrightarrow{} ZnCl_2+H_2↑$

$n_{Zn}=\dfrac{19,5}{65}=0,3mol.$

$Theo$ $pt:$ $n_{H_2}=n_{Zn}=0,3mol.$

$⇒V_{H_2}=0,3.22,4=6,72l.$

$b,PTPƯ:Fe_2O_3+3H_2\xrightarrow{t^o} 2Fe+3H_2O$

$n_{Fe_2O_3}=\dfrac{19,2}{160}=0,12mol.$

$\text{Lập tỉ lệ:}$ $\dfrac{0,12}{1}>\dfrac{0,3}{3}$

$⇒Fe_2O_3$ $dư.$

$Theo$ $pt:$ $n_{Fe}=\dfrac{2}{3}n_{H_2}=0,2mol.$

$⇒m_{Fe}=0,2.56=11,2g.$

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