Câu 1:

$1)\ 3Fe+2O_2\xrightarrow{t^o} Fe_3O_4\\ 2)\ CaO+2HCl\xrightarrow\ CaCl_2+H_2O\\ 3)\ 2Fe(OH)_3\xrightarrow{t^o} Fe_2O_3+3H_2O\\ 4)\ SO_2+2KOH\xrightarrow\ K_2SO_3+H_2O$

Câu 2:

$Mg+2HCl\xrightarrow\ MgCl_2+H_2\\ n_{Mg}=n_{H_2}=\dfrac{2,479}{24,79}=0,1(mol)\\ m_{Mg}=0,1.24=2,4(g)$

Câu 3:

$CaCO_3\xrightarrow{t^o} CaO+CO_2\\ n_{CO_2(lt)}=n_{CaCO_3}=\dfrac{10}{100}=0,1(mol)\\ H=\dfrac{4,48}{0,1.56}.100\%=80\%$

câu 1 $1, 3Fe + 2O_2 \xrightarrow{to} F_3O_4$

$2, CaO + 2HCl \xrightarrow{} CaCl_2 + H_2O$

$3, 2Fe(OH)_3 \xrightarrow{to} Fe_2O_3 +3 H_2O$

$4, SO_2 + 2KOH \xrightarrow{} K_2SO_3 + H_2O$

Câu `2` 

$Mg +2HCl \xrightarrow{} MgCl_2 +H_2$ 

$n_{H_2}=\dfrac{2,479}{24,79}=0,1(mol)$ 

theo pt 

$n_{Mg}=n_{H_2}=0,1(mol)$

$\to m_{Mg}=0,1.24=2,4(g)$  

Câu`3` 

$CaCO_3 \xrightarrow{} CaO +CO_2 $ 

$n_{CaCO_3}=\dfrac{10}{100}=0,1(mol)$ 

theo pt 

$n_{CaO(lt)}=n_{CaCO_3}=0,1(mol)$ 

$\to m_{CaO(lt)}=0,1.56=5,6(g)$ 

`H={4,48}/{5,6}.100%=80%`