Giải thích các bước giải:

$a)$

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

$Zn+H_2SO_4\to ZnSO_4+H_2$

$n_{H_2}=n_{Zn}=0,3(mol)$

$V_{H_2}=0,3.22,4=6,72(l)$

$b)$

$n_{ZnSO_4}=n_{Zn}=0,3(mol)$

$m_{ZnSO_4}=0,3.161=48,3(g)$

$c)$

$n_{FeO}=\dfrac{14,4}{72}=0,2(mol)$

$FeO+H_2\xrightarrow{t^o}Fe+H_2O$

$n_{FeO}<n_{H_2}\to FeO$ dư.

$n_{Fe}=n_{FeO}=0,2(mol)$

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

Đáp án + Giải thích các bước giải:

$a\big)$

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

$Zn+H_2SO_4\to ZnSO_4+H_2$

Theo PT: $n_{H_2}=n_{Zn}=0,3(mol)$

$\to V_{H_2(đktc)}=0,3.22,4=6,72(l)$

$b\big)$

Theo PT: $n_{ZnSO_4}=n_{Zn}=0,3(mol)$

$\to m_{ZnSO_4}=0,3.161=48,3(g)$

$c\big)$

$n_{FeO}=\frac{14,4}{72}=0,2(mol)$

$FeO+H_2\xrightarrow{t^o}Fe+H_2O$

Vì $n_{FeO}<n_{H_2}\to H_2$ dư

Theo PT: $n_{Fe}=n_{FeO}=0,2(mol)$

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