Em tham khảo nha :

\(\begin{array}{l}
1)\\
CaC{O_3} \to CaO + C{O_2}\\
a)\\
{n_{CaC{O_3}}} = \dfrac{m}{M} = \dfrac{{50}}{{100}} = 0,5mol\\
{n_{CaO}} = {n_{CaC{O_3}}} = 0,5mol\\
{m_{CaO}} = n \times M = 0,5 \times 56 = 28g\\
b)\\
{n_{C{O_2}}} = {n_{CaC{O_3}}} = 0,5mol\\
{V_{C{O_2}}} = n \times 22,4 = 0,5 \times 22,4 = 11,2l\\
2)\\
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
a)\\
{n_{F{e_2}{O_3}}} = \dfrac{m}{M} = \dfrac{{16}}{{160}} = 0,1mol\\
{n_{{H_2}}} = 3{n_{F{e_2}{O_3}}} = 0,3mol\\
{V_{{H_2}}} = n \times 22,4 = 0,3 \times 22,4 = 6,72l\\
b)\\
{n_{Fe}} = 2{n_{F{e_2}{O_3}}} = 0,2mol\\
{m_{Fe}} = n \times M = 0,2 \times 56 = 11,2g\\
3)\\
F{e_3}{O_4} + 4{H_2} \to 3Fe + 4{H_2}O\\
a)\\
{n_{{H_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4mol\\
{n_{F{e_3}{O_4}}} = \dfrac{{{n_{{H_2}}}}}{4} = 0,1mol\\
{m_{F{e_3}{O_4}}} = n \times M = 0,1 \times 232 = 23,2g\\
b)\\
C1:{n_{Fe}} = \dfrac{3}{4}{n_{{H_2}}} = 0,3mol\\
{m_{Fe}} = n \times M = 0,3 \times 56 = 16,8g\\
C2:\\
{n_{{H_2}O}} = {n_{{H_2}}} = 0,4mol\\
{m_{{H_2}O}} = n \times M = 0,4 \times 18 = 7,2g\\
\text{Theo định luật bảo toàn khối lượng ta có :}\\
{m_{F{e_3}{O_4}}} + {m_{{H_2}}} = {m_{Fe}} + {m_{{H_2}O}}\\
 \Rightarrow {m_{Fe}} = 23,2 + 0,4 \times 2 - 7,2 = 16,8g
\end{array}\)

$\text{Bài làm:}$

$\text{Câu 1: ta có,}$

$n_{CaCO_3}=\dfrac{m_{CaCO_3}}{M_{CaCO_3}}=\dfrac{50}{100}=0,5\text{(mol)}$

$\text{Phương trình:}$

$CaCO_3 \xrightarrow{t^0} CaO+ CO_2$

$\text{ Ta thấy:}$

$n_{CaCO_3}=n_{CaO}=n_{CO_2}=0,5\text{(mol)}$

$\text{a)} m_{CaO}=56.0,5=28\text{(g)}$

$\text{b)} n_{CO_2}=0,5.22,4= 11,2\text{(l)}$

$\text{Câu 2:}$

$n_{Fe_2O_3}=\dfrac{16}{160}=0,1\text{(mol)}$

$\text{Phương trình:}$

$Fe_2O_3 + 3H_2 \xrightarrow{t^0} 2Fe + 3H_2O$

 $\text{a)} n_{H_2}=0,1.3=0,3\text{(mol)}$

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

$\text{b)} n_{Fe}=0,1.2=0,2\text{(mol)}$

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

$\text{Câu 3:}$

$n_{H_2}=\dfrac{8,96}{22,4}=0,4\text{(mol)}$

$\to m_{H_2}= 0,4.2=0,8\text{(g)}$

$\text{Phương trình:}$

$Fe_3O_4 + 4H_2 \xrightarrow{t^0} 3Fe + 4H_2O$

$\text{a)} n_{Fe_3O_4}=\dfrac{0,4}{4}=0,1\text{(mol)}$

$\to m_{Fe_3O_4}=0,1.232=23,2\text{(g)}$

$\text{b) Cách 1:}$

$n_{Fe}=0,4.\dfrac{3}{4}=0,3\text{(mol)}$

$\to m_{Fe}=56.0,3=16,8\text{(g)}$

$\text{Cách 2:}$

$n_{H_2O}=n_{H_2}=0,4\text{(mol)}$

$\to m_{H_2O}=0,4.18=7,2\text{(g)}$

$\text{Áp dụng định luật bảo toàn khối lượng}$

$\text{Ta có:}$

$m_{Fe_3O_4}+m_{H_2}=m_{Fe}+m_{H_2O}$

$\to m_{Fe}=m_{Fe_3O_4}+m_{H_2}-m_{H_2O}$

$\to m_{Fe}= 23,2+0,8-7,2$

$\to m_{Fe}=16,8\text{(g)}$