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

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

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

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

)

....1................1..................1............1(mol)

0,3................0,3................0,3.........0,3(mol)

Ta có n_H2 = \(\dfrac{6,72}{22,4}\) = 0,3 ( mol )

Fe₃O₄ + 4H₂ \(\rightarrow\) 3Fe + 4H₂O

x................4x.......3x.........4x

ZnO + H₂ \(\rightarrow\) Zn + H₂O

y...........y.........y........y

=> \(\left\{{}\begin{matrix}232x+81y=19,7\\4x+y=0,3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)

a, => m_Fe3O4 = 232 . 0,05 = 11,6 ( gam )

=> m_ZnO = 81 . 0,1 = 8,1 ( gam )

b, => m_Fe = 56 . ( 0,05 . 3 ) = 8,4 ( gam )

=> m_Zn = 65 . 0,1 = 6,5 ( gam )

c,

Mg + H₂SO₄ \(\rightarrow\) MgSO₄ + H₂

0,3........0,3............0,3.........0,3

=> m_Mg = 0,3 . 24 = 7,2 ( gam )

=> m_H2SO4 = 98 . 0,3 = 29,4 ( gam )

=> m_{H2SO4 cần dùng} = 29,4 : 90 . 100 = \(\dfrac{49}{15}\) ( gam )

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%

Gọi số mol Zn, Al là a, b (mol)

=> 65a + 27b = 18,4 (1)

\(n_{O_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

              b-->0,75b

            2Zn + O₂ --t^o--> 2ZnO

              a-->0,5a

=> 0,5a + 0,75b = 0,25 (2)

(1)(2) => a = 0,2 (mol); b = 0,2 (mol)

=> \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,2.65}{18,4}.100\%=70,65\%\\\%m_{Al}=\dfrac{0,2.27}{18,4}.100\%=29,35\%\end{matrix}\right.\)

\(2Zn+O_2\rightarrow 2ZnO \)

\(4Al+3O_2\rightarrow 2Al_2O_3 \)

\(n_{O_2}=\dfrac{5,6}{22,4}=0,25(mol) \)

\(Theo PT : x = 0,2 mol ; y = 0,2 mol \)

\(\%\)\(m_{Zn}=\dfrac{0,2.65}{18,4}.100\)\(\%\)\(=70,65 \)\(\%\)

\(\%\)\(m_{Al}=100\)\(\%\)\(-70,65=29,35\)\(\%\)

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

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

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

Gọi x, y lần lượt là số mol của \(Fe_3O_4,ZnO\)

\(Fe_3O_4+4H_2\rightarrow3Fe+4H_2O\)

1...................4............3............4(mol)

x..................4x.........3x...........4x(mol)

\(ZnO+H_2\rightarrow Zn+H_2O\)

1..............1...........1.........1(mol)

y..............y............y.........y(mol)

Ta có:\(\left\{{}\begin{matrix}232x+81y=19,7\\4x+y=0,3\end{matrix}\right.\)

=>x=0,05

=>y=0.1

\(m_{Fe_3O_4}:232.0,05=11,6\left(g\right)\)

\(m_{ZnO}:19,7-11,6=8,1\left(g\right)\)

b)\(m_{Fe}:56.0,15=8,4\left(g\right)\)

\(m_{Zn}:65.0,1=6,5\left(g\right)\)

c)\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

....1................1..................1............1(mol)

0,3................0,3................0,3.........0,3(mol)

\(m_{Mg}:0,3.24=7,2\left(g\right)\)

\(m_{H_2SO_4}:0,3.98+0,3.98.10\%=32.34\left(g\right)\)

a.

 \(m_{Ag}=m_{k.tan}=8,7\left(g\right)\\ m_{Zn,Mg}=20-8,7=11,3\left(g\right)\\ \left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\\Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ \rightarrow\left\{{}\begin{matrix}65a+24b=11,3\\a+b=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{Ag}=\dfrac{8,7}{20}.100=43,5\%\\\%m_{Mg}=\dfrac{24.0,2}{20}.100=24\%\\\%m_{Zn}=\dfrac{0,1.65}{20}.100=32,5\%\end{matrix}\right.\)

b. 

\(n_{H_2SO_4\left(tổng\right)}=a+b=0,3\left(mol\right)\\ V_{ddH_2SO_4\left(tổng\right)}=\dfrac{0,3}{0,5}=0,6\left(lít\right)=600\left(ml\right)\)