\(m_{Fe_3O_4} = \dfrac{31,2 + 15,2}{2} = 23,2(gam) \Rightarrow n_{Fe_3O_4} = \dfrac{23,2}{232} = 0,1(mol)\\ Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O\\ n_{Fe} = 3n_{Fe_3O_4} = 0,3(mol) \Rightarrow m_{Fe} = 0,3.56 = 16,8(gam)\\ CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ n_{Cu} = n_{CuO} = \dfrac{31,2-23,2}{80} = 0,1(mol) \Rightarrow m_{Cu} = 0,1.64 = 6,4(gam)\)

Ta có:\(\left\{{}\begin{matrix}m_{CuO}+m_{Fe_3O_4}=31,2\\m_{Fe_3O_{\text{4}}}-m_{CuO}=15,2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m_{Fe_3O_4}=23.2\\m_{CuO}=8\end{matrix}\right.\)

\(n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\)

\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

\(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)

\(0,1\rightarrow\)                0.3

\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

\(0,1\rightarrow\)           0,1

\(m_{Fe}=0,3.56=16,8\left(g\right)\)

\(m_{Cu}=0,1.64=6,4\left(g\right)\)

dùng tổng hiệu tính ra m CuO và m Fe3O4

m CuO = (31,2-15,2):2

m Fe3O4= 31,2-m CuO

viết PTHH của từng cái tác dụng vs H2 => tính ra mol của Fe và mol H2 => tính khối lượng Fe3O4 và khối lượng CuO => cộng lại

mình nghĩ vậy thôi... sai thì sorry nha

\(\left\{\begin{matrix}m_{Fe_3O_4}+m_{CuO}=31,2\left(g\right)\\m_{Fe_3O_4-}m_{CuO}=15,2\left(g\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{\begin{matrix}m_{Fe_3O_4}=23,2\left(g\right)\\m_{CuO}=8\left(g\right)\end{matrix}\right.\)

\(\Rightarrow\)\(n_{Fe_3O_4}=0,1\left(mol\right)\)

\(\Rightarrow n_{CuO}=0.1\left(mol\right)\)

PTPƯ : CuO + \(H_2\) \(\underrightarrow{t^o}\) Cu + \(H_2O\) (1)

\(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\) (2)

Theo PTPƯ \(n_{Cu}=n_{CuO}=0,1\left(mol\right)\)

\(\Rightarrow m_{Cu}=64.0,1=6,4\left(g\right)\)

- \(3n_{Fe}=n_{Fe_3O_4}=0,1.3=0.3\left(mol\right)\)

\(\Rightarrow m_{Fe}=56.0,3=16,8\left(g\right)\)

Gọi khối lượng của CuO và Fe₃O₄ lần lược là x, y thì ta có hệ:

\(\left\{\begin{matrix}x+y=31,2\\y-x=15,2\end{matrix}\right.\)

\(\Leftrightarrow\left\{\begin{matrix}x=8\\y=23,2\end{matrix}\right.\)

\(CuO\left(0,1\right)+H_2\rightarrow Cu\left(0,1\right)+H_2O\)

\(Fe_3O_4\left(0,1\right)+4H_2\rightarrow3Fe\left(0,3\right)+4H_2O\)

\(n_{CuO}=\frac{8}{80}=0,1\left(mol\right)\)

\(\Rightarrow m_{Cu}=0,1.64=6,4\left(g\right)\)

\(n_{Fe_3O_4}=\frac{23,2}{232}=0,1\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,3.56=16,8\left(g\right)\)

a) \(m_{CuO}=\dfrac{20.40}{100}=8\left(g\right)\) => \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

\(m_{Fe_2O_3}=20-8=12\left(g\right)\) => \(n_{Fe_2O_3}=\dfrac{12}{160}=0,075\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

             0,1--->0,1------>0,1

            Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

          0,075--->0,225----->0,15

=> m_Cu = 0,1.64 = 6,4 (g)

=> m_Fe = 0,15.56 = 8,4 (g)

b) \(V_{H_2}=\left(0,1+0,225\right).22,4=7,28\left(l\right)\)

\(Tacó:m_{Fe_2O_3}:m_{CuO}=3:2\\ m_{Fe_2O_3}+m_{CuO}=40\\ \Rightarrow m_{Fe_2O_3}=24\left(g\right)\Rightarrow n_{Fe_2O_3}=0,15\left(mol\right)\\ \Rightarrow m_{CuO}=16\left(g\right)\Rightarrow n_{CuO}=0,2\left(mol\right)\\ Fe_2O_3+3H_2-^{t^o}\rightarrow2Fe+3H_2O\left(1\right)\\ CuO+H_2O-^{t^o}\rightarrow Cu+H_2O\left(2\right)\\TheoPT\left(1\right): n_{Fe}=2n_{Fe_2O_3}=0,3\left(mol\right)\\ \Rightarrow m_{Fe}=16,8\left(g\right)\\TheoPT\left(2\right): n_{Cu}=n_{CuO}=0,2\left(mol\right)\\ \Rightarrow m_{Cu}=0,2.64=12,8\left(g\right)\)

\(a)Gọi : n_{CuO} = x(mol) \Rightarrow n_{Fe_2O_3} = \dfrac{80a.2}{160}=x(mol)\\ CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + H_2O\\ n_{H_2} = x + x = \dfrac{8,96}{22,4} = 0,4(mol)\Rightarrow x = 0,2\\ a = 0,2.80 + 0,2.160 = 48(gam)\\ b)\\ n_{Fe} = 2n_{Fe_2O_3} = 0,4(mol) \Rightarrow m_{Fe} = 0,4.56 = 22,4(gam)\\ n_{Cu} = n_{CuO} = 0,2(mol) \Rightarrow m_{Cu} = 0,2.64 = 12,8(gam)\)

\(Fe_2O_3\left(0,075\right)+3H_2\rightarrow2Fe\left(0,15\right)+3H_2O\)

\(CuO\left(0,1\right)+H_2\rightarrow Cu\left(0,1\right)+H_2O\)

\(m_{Fe_2O_3}=20.60\%=12\left(g\right)\)

\(\Rightarrow n_{Fe_2O_3}=\dfrac{12}{160}=0,075\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)

\(m_{CuO}=20.40\%=8\left(g\right)\)

\(\Rightarrow n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

\(\Rightarrow m_{Cu}=0,1.64=6,4\left(g\right)\)