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

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

Bài 37 :

PTHH :

* Dẫn A qua nước vôi trong

CO₂ + Ca(OH)₂ -> CaCO₃ \(\downarrow\) + H₂O (1)

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

* Đốt cháy A rồi dẫn sản phẩm qua nước vôi trong (2)

2CO + O₂ -> 2CO₂

x......................x (mol)

CO₂ + Ca(OH)₂ -> CaCO₃ \(\downarrow\) + H₂O

x.................................x (mol)

CO₂ + Ca(OH)₂ -> CaCO₃ \(\downarrow\) + H₂O

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

Gọi x,y lần lượt là số mol CO và CO₂ trong hỗn hợp (x,y>0)

n_CaCO3(1) = \(\dfrac{30}{100}=0,3\left(mol\right)\) => y=0,3 (mol)

\(\Sigma n_{CaCO3\left(2\right)}=x+y=\dfrac{50}{100}=0,5\left(mol\right)\)

=> x = 0,2 (mol)

=> \(\left\{{}\begin{matrix}m_{CO}=0,2.28=5,6\left(g\right)\\m_{CO2}=0,3.44=13,2\left(g\right)\end{matrix}\right.\)

Vì đo ở cùng điều kiện nên tỉ lệ thể tích bằng ỉ lệ số mol

n_A = x+y = 0,5 (mol)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CO}=\dfrac{0,2}{0,5}.100\%=40\%\\\%V_{CO2}=\dfrac{0,3}{0,5}.100\%=60\%\end{matrix}\right.\)

a,PTHH: C   +  O2  ->  CO2   (*)

            x        x             x    (mol)

           S  +  O2  ->  SO2     (**)

           y       y            y      (mol)

Ta có dB/H2=9 => MB = 29.2=58 (g)=m hh khí B / n hh Khí B

<=>58= \(\frac{44x+64y}{x+y}\)

=>44x+64y=58x+58y

<=>7x=3y

<=> 7x-3y=0   (1)

Mà hh A =12x+32y=13 (2)

b,Từ (1),(2) ta có hệ pt

\(\hept{\begin{cases}7x-3y=0\\12x+3y=13\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0,15\\y=0,35\end{cases}}\)

=>%C = (12.0,15)/13 .100%=13,84%

    %S= 100%-13,84%= 86,16%

Từ pt (*),(**) ta có VO2(đktc)=(x+y).22,4=11,2(l)

\(n_{Mg}=2x\left(mol\right),n_{Fe}=x\left(mol\right)\)

\(n_{HCl}=0.2\cdot0.45=0.9\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(n_{HCl}=2\cdot2x+2\cdot x=0.9\left(mol\right)\)

\(\Rightarrow x=0.15\)

\(m_{hh}=0.3\cdot24+0.15\cdot56=15.6\left(g\right)\)

\(V_{H_2}=0.45\cdot22.4=10.08\left(l\right)\)

Bài 1 : 

\(a) Fe_2O_3 + 3H_2 \xrightarrow{t^o}2Fe + 3H_2O\\ b) n_{Fe_2O_3} = \dfrac{80}{160}= 0,5(mol)\\ n_{H_2} = 3n_{Fe_2O_3} = 1,5(mol)\\ \Rightarrow V_{H_2} = 1,5.22,4 = 33,6(lít)\\ n_{Fe} = 2n_{Fe_2O_3} = 1(mol)\\ m_{Fe} = 1.56 = 56(gam)\)

Bài 2 :

\(a) Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} = n_{Fe} =\dfrac{5,6}{56} = 0,1(mol)\\ V_{H_2} = 0,1.22,4 = 2,24(lít)\\ n_{HCl} =2 n_{Fe} = 0,2(mol)\\ m_{HCl} = 0,2.36,5 = 7,3(gam)\)

\(n_{Fe_3O_4}=\dfrac{4,64}{232}=0,02mol\)

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

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

0,02        0,4          0         0

0,02        0,08       0,06    0,08

0             0,32       0,06    0,08

b)\(m_{Fe}=0,06\cdot56=3,36g\)   

\(m_{H_2O}=0,08\cdot18=1,44g\)

c)\(Fe+2HCl\rightarrow FeCl_2+H_2\)

   0,06     0,5      0            0

\(\Rightarrow\)Tính theo \(Fe\)

\(\Rightarrow V_{H_2}=0,06\cdot22,4=1,344l\)