a, mFe2O3 = 32 . 75% = 24 (g)

nFe2O3 = 24/160 = 0,15 (mol)

mCuO = 32 - 24 = 8 (g)

nCuO = 8/80 = 0,1 (mol)

PTHH:

Fe2O3 + 3H2 -> (t°) 2Fe + 3H2O

0,15 ---> 0,45 ---> 0,3 

CuO + H2 -> (t°) Cu + H2O

0,1 ---> 0,1 ---> 0,1

mFe = 0,3 . 56 = 16,8 (g)

mCu = 64 . 0,1 = 6,4 (g)

b, nH2 = 0,1 + 0,45 = 0,55 (mol)

VH2 = 0,55 . 22,4 = 12,32 (l)

c, PTHH:

2Al + 6HCl -> 2AlCl3 + 3H2

11/30 <--- 1,1 <--- 11/30 <--- 0,55

mAl = 11/30 . 27 = 9,9 (g)

mHCl = 1,1 . 36,5 = 40,15 (g)

\(a) n_{Fe_2O_3} = \dfrac{16.75\%}{160} = 0,075(mol)\\ n_{CuO} = \dfrac{16.25\%}{80} = 0,05(mol)\\ Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O\\ CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ n_{Fe} = 2n_{Fe_2O_3} = 0,15(mol) \Rightarrow m_{Fe} = 0,15.56 = 8,4(gam)\\ n_{Cu} = n_{CuO} = 0,05(mol) \Rightarrow m_{Cu} = 0,05.64 = 3,2(gam)\\ b) n_{H_2} = 3n_{Fe_2O_3} + n_{CuO} = 0,075.3 + 0,05 = 0,275(mol)\\ V_{H_2} = 0,275.22,4 = 6,16(lít)\)

\(m_{Fe_2O_3}=16\cdot75\%=12\left(g\right)\)

\(n_{Fe_2O_3}=\dfrac{12}{160}=0.075\left(mol\right)\)

\(n_{CuO}=16\cdot25\%=4\left(g\right)\)

\(n_{CuO}=\dfrac{4}{80}=0.05\left(mol\right)\)

\(Fe_2O_3+3H_2\underrightarrow{^{^{t^0}}}2Fe+3H_2O\)

\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)

\(n_{H_2}=3\cdot0.075+0.05=0.275\left(mol\right)\)

a,\(m_{Fe_2O_3}=16.75\%=12\left(g\right)\Rightarrow n_{Fe_2O_3}=\dfrac{12}{160}=0,075\left(mol\right)\)

   \(m_{CuO}=16-12=4\left(g\right)\Rightarrow n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂ → 2Fe + 3H₂O

Mol:     0,075    0,225    0,15

PTHH: CuO + H₂ → Cu + H₂O

Mol:     0,05   0,05   0,05

\(\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right);m_{Cu}=0,05.64=3,2\left(g\right)\)

b,\(n_{H_2}=0,225+0,05=0,275\left(mol\right)\)

\(n_{Pb}=\dfrac{2,07}{207}=0,01mol\)

\(n_{Cu}=\dfrac{1,6}{64}=0,025mol\)

\(PbO+H_2\rightarrow\left(t^o\right)Pb+H_2O\)

0,01    0,01            0,01            ( mol )

\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)

0,025  0,025        0,025            ( mol )

\(m_{hh}=m_{PbO}+m_{CuO}=\left(0,01.223\right)+\left(0,025.80\right)=4,23g\)

\(V_{H_2}=\left(0,01+0,025\right).22,4=0,784l\)

\(n_{Pb}=\dfrac{2,07}{207}=0,01\left(mol\right)\\ n_{Cu}=\dfrac{1,6}{64}=0,025\left(mol\right)\\ PTHH:PbO+H_2\underrightarrow{t^o}Pb+H_2O\\ Mol:0,01\leftarrow0,01\leftarrow0,01\\ CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ Mol:0,025\leftarrow0,025\leftarrow0,025\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CuO}=0,025.80=2\left(g\right)\\m_{PbO}=0,01.223=2,23\left(g\right)\end{matrix}\right.\Rightarrow m_{oxit}=2+2,23=4,23\left(g\right)\\ V_{H_2}=\left(0,01+0,025\right).22,4=0,784\left(l\right)\)

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