Đáp án + Giải thích các bước giải:

$a)$

$n_{Zn}=\frac{9,75}{65}=0,15(mol)$

$Zn+2HCl\to ZnCl_2+H_2$

Theo PT: $n_{ZnCl_2}=n_{Zn}=0,15(mol)$

$\to m_{ZnCl_2}=0,15.136=20,4(g)$

$b)$

Theo PT: $n_{HCl}=2n_{Zn}=0,3(mol)$

$\to m_{HCl}=0,3.36,5=10,95(g)$

$c)$

Theo PT: $n_{H_2}=n_{Zn}=0,15(mol)$

$Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O$

Theo PT: $n_{Fe_2O_3}=\frac{1}{3}n_{H_2}=0,05(mol)$

$\to m_{Fe_2O_3}=0,05.160=8(g)$

Theo PT: $n_{Fe}=\frac{2}{3}n_{H_2}=0,1(mol)$

$\to m_{Fe}=0,1.56=5,6(g)$

Đáp án+ Giải thích các bước giải:

$a)$

$Zn + 2HCl$ $\rightarrow$ $ZnCl_2 + H_2$

Theo pt: $n_{ZnCl_2}$= $n_{H_2}$=  $n_{Zn}$= $\frac{9,75}{65}$= $0,15(mol)$

$\rightarrow$  $m_{ZnCl_2}$= $0,15.136=20,4(g)$

$b)$

Theo pt: $n_{HCl}$= $2n_{Zn}$= $0,15.2=0,3(mol)$

$\rightarrow$ $m_{HCl}$= $0,3.36,5= 10,95(g)$

$c)$

$Fe_2O_3 + 3H_2$ $\xrightarrow{t^o}$ $2Fe + 3H_2O$

Theo pt: $n_{Fe_2O_3}$= $\frac{0,15.1}{3}$ = $0,05(mol)$

$\rightarrow$ $m_{Fe_2O_3}$= $0,05.160=8(g)$

Theo pt: $n_{Fe}$= $\frac{0,15.2}{3}$ = $0,1(mol)$

$\rightarrow$ $m_{Fe}$= $0,1.56=5,6(g)$