2. \(1.n_{H_2}=0,1\left(mol\right);n_{CuO}=0,15\left(mol\right)\\ H_2+CuO-^{t^o}\rightarrow Cu+H_2O\\ LTL:\dfrac{0,1}{1}< \dfrac{0,15}{1}\Rightarrow CuOdư\\ n_{CuO\left(pư\right)}=n_{H_2}=0,1\left(mol\right)\\ n_{CuO\left(dư\right)}=0,15-0,1=0,05\left(mol\right)\\ \Rightarrow m_{CuO\left(dư\right)}=4\left(g\right)\\ 2.n_{H_@}=n_{Cu}=0,1\left(mol\right)\\ \Rightarrow m_{Cu}=0,1.64=6,4\left(g\right)\)

Bài 9:

Gọi hóa trị của M là x(x>0)

\(n_{H_2}=\dfrac{1,008}{22,4}=0,045(mol)\\ \text {Bảo toàn O: }n_{O/oxit}=n_{H_2}=\dfrac{1,344}{22,4}=0,06(mol)\\ \Rightarrow m_{M}=m_{oxit}-m_{O_2}=3,48-0,06.16=2,52(g)\\ PTHH:2M+2xHCl\to 2MCl_x+xH_2\\ \Rightarrow n_{M}=\dfrac{2.0,045}{x}=\dfrac{0,09}{x}\\ \Rightarrow M_{M}=\dfrac{2,52}{\dfrac{0,09}{x}}=28x\)

Thay \(x=2\Rightarrow M_M=56(Fe)\)

\(\Rightarrow n_{Fe}=\dfrac{2,52}{56}=0,045(mol)\\ \Rightarrow \dfrac{n_{Fe}}{n_{O}}=\dfrac{0,045}{0,06}=\dfrac{3}{4}\)

Vậy M là sắt (Fe) và CTHH oxit là \(Fe_3O_4\)

9. Tham khảo

\(n_{H_2}=\dfrac{4,032}{22,4}=0,18\left(mol\right)\)

PTHH: Fe + H₂SO₄ ---> FeSO₄ + H₂

            0,18 <------------------------ 0,18

\(\rightarrow n_O=\dfrac{13,92-0,18.56}{16}=0,24\left(mol\right)\)

CTHH: Fe_xO_y

=> x : y = 0,18 : 0,24 = 3 : 4

CTHH Fe₃O₄

PTHH: \(Fe_xO_y+yH_2\underrightarrow{t^o}xFe+yH_2O\)

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

Đặt \(\left\{{}\begin{matrix}n_{Fe\left(oxit\right)}=a\left(mol\right)=n_{H_2}\\n_{O\left(oxit\right)}=b\left(mol\right)\end{matrix}\right.\)

Ta có: \(m_{tăng}=m_{Fe}-m_{H_2}\) \(\Rightarrow56a-2a=3,24\) \(\Rightarrow a=n_{Fe}=0,06\left(mol\right)\)

Hỗn hợp D gồm \(\left\{{}\begin{matrix}n_{CO_2\left(dư\right)}=c\left(mol\right)\\n_{H_2O}=n_{O\left(oxit\right)}=b\left(mol\right)\end{matrix}\right.\)

Ta có hệ phương trình: \(\left\{{}\begin{matrix}c+b=0,1\\18b+2c=7,4\cdot2\cdot\left(b+c\right)\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=0,08\\c=0,02\end{matrix}\right.\)

\(\Rightarrow x:y=a:b=0,06:0,08=3:4\)

\(\Rightarrow\)  Công thức cần tìm là Fe₃O₄ 

\(V_{H_2\left(đktc\right)}=\dfrac{4,032}{22,4}=0,18\left(mol\right)\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

1                                     :   1   (mol)

0,18                                :  0,18 (mol)

\(yCO+Fe_xO_y\rightarrow^{t^0}xFe+yCO_2\uparrow\)

               1      :         x   (mol)

              \(\dfrac{0,18}{x}\)                0,18  (mol)

\(M_{Fe_xO_y}=\dfrac{m}{n}=\dfrac{13,92}{\dfrac{0,18}{x}}=\dfrac{232}{3}x\)

\(\Rightarrow56x+16y=\dfrac{232}{3}x\)

\(\Rightarrow16y=\dfrac{64}{3}x\)

\(\Rightarrow\dfrac{x}{y}=\dfrac{16}{\dfrac{64}{3}}=\dfrac{3}{4}\Rightarrow x=3;y=4\)

-Vậy CTHH của oxit sắt là Fe₃O₄

- Hỗn hợp rắn sau pư gồm: Fe và FeCl₃.

Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

BT e, có: \(2n_{Fe}=2n_{H_2}\Rightarrow n_{Fe}=0,1\left(mol\right)\)

BTNT Fe, có: \(n_{FeCl_3}=n_{Fe\left(OH\right)_3}=0,03\left(mol\right)\) = n_{Fe (pư)}

BTNT Fe, có: n_{Fe (ban đầu)} = n_Fe + n_FeCl3 = 0,13 (mol)

\(\Rightarrow H\%=\dfrac{0,03}{0,13}.100\%\approx23,077\%\)

\(a,PTHH:Fe_3O_4+4H_2\xrightarrow{t^o}3Fe+4H_2O\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol);n_{Fe_3O_4}=\dfrac{46,4}{232}=0,2(mol)\)

Vì \(\dfrac{n_{H_2}}{4}<\dfrac{n_{Fe_3O_4}}{1}\) nên \(Fe_3O_4\) dư

\(n_{Fe_3O_4(dư)}=0,2-\dfrac{0,3}{4}=0,125(mol)\\ \Rightarrow m_{Fe_3O_4(dư)}=0,125.232=29(g)\\ b,n_{Fe}=\dfrac{3}{4}n_{H_2}=0,225(mol)\\ \Rightarrow m_{Fe}=0,225.56=12,6(g)\)

\(a,PTHH:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\\ n_{Al}=\dfrac{m}{M}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ Theo.PTHH:n_{Al_2O_3}=\dfrac{1}{2}.n_{Al}=\dfrac{1}{2}.0,4=0,2\left(mol\right)\\ m_{Al_2O_3}=n.M=0,2.102=20,4\left(g\right)\)

\(b,n_{O_2}=\dfrac{V_{\left(đktc\right)}}{22,4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ Lập.tỉ.lệ:\dfrac{n_{Al}}{4}>\dfrac{n_{O_2}}{3}\Rightarrow Al.dư\\ Theo.PTHH:n_{Al\left(pư\right)}=\dfrac{4}{3}.n_{O_2}=\dfrac{4}{3}.0,2\left(mol\right)\\ n_{Al\left(dư\right)}=n_{Al\left(bđ\right)}-n_{Al\left(pư\right)}=0,4-0,2=0,2\left(mol\right)\\ Theo.PTHH:n_{Al_2O_3}=\dfrac{1}{2}.n_{Al}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\ m_{Al_2O_3}=n.M=0,1=102=10,2\left(g\right)\)

- Cho phản ứng xảy ra hoàn toàn (2 chất trong A có sắt và oxit khác oxit sắt ban đầu)

\(yH_2+Fe_xO_y\rightarrow\left(t^o\right)xFe+yH_2O\left(1\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\left(2\right)\\ n_{H_2\left(2\right)}=n_{Fe\left(2\right)}=n_{Fe\left(1\right)}=0,3\left(mol\right)\\ n_{O\left(trong.oxit\right)}=n_{H_2O}=n_{H_2}=0,4\left(mol\right)\\ BTKL:m_{H_2}+m_{oxit}=m_A+m_{H_2O}\\ \Leftrightarrow0,4.2+m=28,4+18.0,4\\ \Leftrightarrow m=34,8\left(g\right)\\ b,x:y=0,3:0,4=3:4\Rightarrow x=3;y=4\\ \Rightarrow CTHH:Fe_3O_4\)