Bài 1:

a) PTHH: \(Fe_2O_3+3H_2\xrightarrow[]{t^o}2Fe+2H_2O\)

                \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

b) Ta có: \(\left\{{}\begin{matrix}m_{Fe_2O_3}=20\cdot80\%=16\left(g\right)\\m_{CuO}=20-16=4\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\\n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{H_2}=3n_{Fe_2O_3}+n_{CuO}=0,35\left(mol\right)\) \(\Rightarrow V_{H_2}=0,35\cdot22,4=7,84\left(l\right)\)

c) Theo các PTHH: \(\left\{{}\begin{matrix}n_{Fe}=2n_{Fe_2O_3}=0,2\left(mol\right)\\n_{Cu}=n_{CuO}=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{hhB}=m_{Fe}+m_{Cu}=0,2\cdot56+0,05\cdot64=14,4\left(g\right)\)

Bài 2:

PTHH: \(Fe_2O_3+3H_2\xrightarrow[]{t^o}2Fe+3H_2O\)

            \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

a) Vì khối lượng Cu bằng \(\dfrac{6}{5}\) khối lượng Fe 

\(\Rightarrow\left\{{}\begin{matrix}m_{Cu}=\dfrac{26,4}{6+5}\cdot6=14,4\left(g\right)\\m_{Fe}=26,4-14,4=12\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{Cu}=\dfrac{14,4}{64}=0,225\left(mol\right)\\n_{Fe}=\dfrac{12}{56}=\dfrac{3}{14}\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{H_2}=\dfrac{3}{2}n_{Fe}+n_{Cu}=\dfrac{9}{28}+0,225=\dfrac{153}{280}\left(mol\right)\) \(\Rightarrow V_{H_2}=\dfrac{153}{280}\cdot22,4=12,24\left(l\right)\)

b) Theo các PTHH: \(\left\{{}\begin{matrix}n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=\dfrac{3}{28}\left(mol\right)\\n_{CuO}=n_{Cu}=0,225\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Fe_2O_3}=\dfrac{3}{28}\cdot160\approx17,14\left(g\right)\\m_{CuO}=0,225\cdot80=18\left(g\right)\end{matrix}\right.\) \(\Rightarrow m_{hh}=35,14\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe_2O_3}=\dfrac{17,14}{35,14}\cdot100\%\approx48,78\%\\\%m_{CuO}=51,22\%\end{matrix}\right.\)

Bài 3:

PTHH: \(CuO+CO\xrightarrow[]{t^o}Cu+CO_2\uparrow\)

            \(Fe_3O_4+4CO\xrightarrow[]{t^o}3Fe+4CO_2\uparrow\)

a) Ta có: \(\left\{{}\begin{matrix}m_{Fe_3O_4}=\dfrac{31,2+15,2}{2}=23,2\left(g\right)\\m_{CuO}=31,2-23,2=8\left(g\right)\end{matrix}\right.\)

\(\Rightarrow n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)=\dfrac{8}{80}=n_{CuO}\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=0,3\left(mol\right)\\n_{Cu}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,3\cdot56=16,8\left(g\right)\\m_{Cu}=0,1\cdot64=6,4\left(g\right)\end{matrix}\right.\)

b) Theo các PTHH: \(n_{CO}=n_{CO_2}=n_{Cu}+\dfrac{4}{3}n_{Fe}=\dfrac{7}{30}\left(mol\right)\)

\(\Rightarrow V_{CO}=V_{CO_2}=\dfrac{7}{30}\cdot22,4\approx5,23\left(l\right)\)  

Tính %m mỗi oxit chứ:v

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Gọi \(\left\{{}\begin{matrix}n_{Fe_3O_4}=x\left(mol\right)\\n_{ZnO}=y\left(mol\right)\end{matrix}\right.\)

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

 x --------> 4x ---> 3x

\(ZnO+H_2\underrightarrow{t^o}Zn+H_2O\)

y ------> y --> y

Có hệ phương trình \(\left\{{}\begin{matrix}232x+81y=19,7\\4x+y=0,3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)

\(\%_{m_{Fe_3O_4}}=\dfrac{232.0,05.100}{19,7}=58,88\%\)

\(\%_{m_{ZnO}}=\dfrac{81.0,1.100}{19,7}=41,12\%\)

\(n_{Fe}=3x=3.0,05=0,15\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\\ n_{Zn}=y=0,1\Rightarrow m_{Zn}=0,1.65=6,5\left(g\right)\)

a) CuO + H₂ -> Cu + H₂O (1)

ZnO + H₂ -> Zn + H₂O (2)

n_{\(_{H_2}\)} = \(\dfrac{4,48}{22,4}\) = 0,2 (mol) => m_{\(H_2\)} = 0,2.2 = 0,4 (g)

Theo PT (1) và (2) ta có: n_{\(H_2O\)} = n_{\(H_2\)} = 0,2 (mol)

=> m_{\(H_2O\)} = 0,2.18 = 3,6 (g)

Theo định luật bảo toàn khối lượng ta có:

m_hh + m_{\(H_2\)} = m_{kim loại} + m_{\(H_2O\)}

=> m_hh = 12,9 + 3,6 - 0,4 = 16,1 (g)

b) Gọi m_Cu là x(g) (0<x<12,9) => n_Cu = \(\dfrac{x}{64}\) (mol)

Thì m_Zn là 12,9-x (g) => n_Zn = \(\dfrac{12,9-x}{65}\) (mol)

Theo PT (1) ta có: n_{\(H_2\)} = n_Cu = \(\dfrac{x}{64}\) (mol)

Theo PT (2) ta có: n_{\(H_2\)} = n_Zn = \(\dfrac{12,9-x}{65}\) (mol)

Theo đề bài, n_{\(H_2\)} là 0,2mol nên ta có:

\(\dfrac{x}{64}+\dfrac{12,9-x}{65}=0,2\) <=> x = 6,4 (g)

=> m_Cu = 6,4 (g)

Vậy: %Cu = \(\dfrac{6,4}{12,9}\).100% \(\approx\) 49,61%

%Zn = 100% - 49,61% = 50,39%

\(n_{Ca\left(OH\right)_2}=0,2.2,5=0,5\left(mol\right)\)

\(CuO+CO\underrightarrow{t^o}Cu+CO_2\uparrow\)

\(Fe_2O_3+CO\underrightarrow{t^o}2FeO+CO_2\uparrow\)

\(CO+PbO\underrightarrow{t^o}CO_2\uparrow+Pb\)

\(FeO+CO\underrightarrow{t^o}Fe+CO_2\uparrow\)

--> Hh kim loại Y là Cu , Pb , FeO , Fe 

hh khí Z sau p/u là CO2 

                          \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\)

trc p/u :                            0,5                  0,3 

p/u :                     0,3          0,3           0,3            0,3 

sau :                    0,3         0,2              0              0,3 

\(n_{CaCO_3}=\dfrac{30}{100}=0,3\left(mol\right)\)

\(m_{CO_2}=0,3.44=13,2\left(g\right)\)

Áp dụng định luật bảo toàn khổi lượng :

\(m_Y=53,5-13,2=40,3\left(g\right)\)

\(Đặt:n_{CuO}=a\left(mol\right);n_{PbO}=b\left(mol\right)\left(a,b>0\right)\\ n_{H_2O}=\dfrac{1,35}{18}=0,075\left(mol\right)\\ CuO+H_2\underrightarrow{^{to}}Cu+H_2O\\ PbO+H_2\underrightarrow{^{to}}Pb+H_2O\\ \Rightarrow\left\{{}\begin{matrix}80a+223b=8,145\\a+b=0,075\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,06\\b=0,015\end{matrix}\right.\\ \Rightarrow\%m_{CuO}=\dfrac{0,06.80}{8,145}.100\approx58,932\%\\ \Rightarrow\%_{PbO}\approx41,068\%\)

Đổi 2,016 dm3 = 2,016 l

nH2 = 2,016/22,4 = 0,09 (mol)

Gọi nFe2O3 = a (mol); nCuO = b (mol)

160a + 80b = 5,6 (g) (1)

PTHH:

Fe2O3 + 3H2 -> (t°) 2Fe + 3H2O

Mol: a ---> 3a ---> 2a ---> 3a

CuO + H2 -> (t°) Cu + H2O

Mol: b ---> b ---> b ---> b

3a + b = 0,09 (mol) (2)

Từ (1) và (2) => a = 0,02 (mol); b = 0,03 (mol)

mFe2O3 = 0,02 . 160 = 3,2 (g)

mCuO = 0,03 . 80 = 2,4 (g)

mH2O = (0,02 . 3 + 0,03) . 18 = 1,62 (g)

mFe = 2 . 0,02 . 56 = 2,24 (g)

mCu = 0,03 . 64 = 1,92 (g)