Kim loại cần tìm đặt là A.

=> CTHH oxit: A₂O₃

\(A_2O_3+H_2SO_4\rightarrow A_2\left(SO_4\right)_3+3H_2O\\ m_{ddsau}=10,2+331,8=342\left(g\right)\\ m_{A_2\left(SO_4\right)_3}=\dfrac{342}{100}.10=34,2\left(g\right)\\ n_{oxit}=\dfrac{34,2-10,2}{96.3-16.3}=0,1\left(mol\right)\\ M_{A_2O_3}=\dfrac{10,2}{0,1}=102\left(\dfrac{g}{mol}\right)=2M_A+48\left(\dfrac{g}{mol}\right)\\ \Rightarrow M_A=\dfrac{102-48}{2}=27\left(\dfrac{g}{mol}\right)\\ \Rightarrow A:Nhôm\left(Al=27\right)\\ \Rightarrow CTHH.oxit:Al_2O_3\)

- Trích mẫu thử.

- Hòa tan từng mẫu thử vào nước có quỳ tím.

+ Không tan: SiO₂.

+ Tan, quỳ tím hóa đỏ: P₂O₅.
PT: \(P_2O_5+3H_2O\rightarrow2H_3PO_4\)

+ Tan, quỳ tím hóa xanh: Na₂O, BaO. (1)

PT: \(Na_2O+H_2O\rightarrow2NaOH\)

\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)

- Cho dd thu được từ mẫu thử nhóm (1) pư với dd H₃PO₄ ở trên.

+ Xuất hiện kết tủa: BaO.
PT: \(3Ba\left(OH\right)_2+2H_3PO_4\rightarrow Ba_3\left(PO_4\right)_2+6H_2O\)

+ Không hiện tượng: Na₂O.

PT: \(3NaOH+H_2PO_4\rightarrow Na_3PO_4+3H_2O\)

- Dán nhãn.

- Trích mẫu thử.

- Cho từng mẫu thử hỗn hợp pư với dd NaOH.

+ Tan 1 phần: Al₂O₃ và Fe₂O₃, Fe và Al₂O₃. (1)

PT: \(Al_2O_3+2NaOH\rightarrow2NaAlO_2+H_2O\)

+ Không tan: Fe và Fe₂O₃.

- Cho mẫu thử nhóm (1) pư với dd HCl.

+ Tan, có khí thoát ra: Fe và Al₂O₃.

+ Tan: Al₂O₃ và Fe₂O₃.

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

- Dán nhãn.

Dd B chứa NaOH.

PT: \(NaOH+HCl\rightarrow NaCl+H_2O\)

Ta có: \(n_{NaCl}=\dfrac{4,68}{58,5}=0,08\left(mol\right)\)

Theo PT: \(n_{NaOH\left(80\left(g\right)dd\right)}=n_{NaCl}=0,08\left(mol\right)\)

\(\Rightarrow n_{NaOH\left(200\left(g\right)dd\right)}=\dfrac{0,08.200}{80}=0,2\left(mol\right)\)

PT: \(2Na+2H_2O\rightarrow2NaOH+H_2\)

\(Na_2O+H_2O\rightarrow2NaOH\)

Ta có: \(\left\{{}\begin{matrix}23n_{Na}+62n_{Na_2O}=5,4\\n_{Na}+2n_{Na_2O}=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_{Na}=0,1\left(mol\right)\\n_{Na_2O}=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{H_2}=\dfrac{1}{2}n_{Na}=0,05\left(mol\right)\)

Ta có: m _{dd B} = m_A + m_H2O - m_H2

⇒ 200 = 5,4 + m_H2O - 0,05.2

⇒ m_H2O = 194,7 (g)

Ta có: \(n_{CO_2}=n_{CaCO_3}=\dfrac{1}{100}=0,01\left(mol\right)\)

\(n_A=0,01\left(mol\right)\)

\(\Rightarrow x=\dfrac{n_{CO_2}}{n_A}=1\)

Vậy: CTPT của A là CH₄.

a, PT: \(4M+3O_2\underrightarrow{t^o}2M_2O_3\)

Ta có: \(n_M=\dfrac{10,8}{M_M}\left(mol\right)\)

\(n_{M_2O_3}=\dfrac{20,4}{2M_M+16.3}\left(mol\right)\)

Theo PT: \(n_M=2n_{M_2O_3}\Rightarrow\dfrac{10,8}{M_M}=2.\dfrac{20,4}{2M_M+16.3}\)

\(\Rightarrow M_M=27\left(g/mol\right)\)

→ M là Nhôm (Al)

b, Ta có: \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,3\left(mol\right)\) \(\Rightarrow V_{O_2}=0,3.22,4=6,72\left(l\right)\)

\(\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=33,6\left(l\right)\)

c, PT: \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

\(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,2\left(mol\right)\)

\(n_{HCl}=6n_{Al_2O_3}=1,2\left(mol\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{1,2}{2}=0,6\left(l\right)\)

d, PT: \(Al_2O_3+2NaOH\rightarrow2NaAlO_2+H_2O\)

Theo PT: \(n_{NaOH}=2n_{Al_2O_3}=0,4\left(mol\right)\)

\(\Rightarrow m_{NaOH}=0,4.40=16\left(g\right)\Rightarrow m_{ddNaOH}=\dfrac{16}{25\%}=64\left(g\right)\)

\(\Rightarrow V_{ddNaOH}=\dfrac{64}{1,25}=51,2\left(ml\right)\)

a, PT: \(R_2O_3+3H_2SO_4\rightarrow R_2\left(SO_4\right)_3+3H_2O\)

Ta có: \(m_{H_2SO_4}=168.35\%=58,8\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{58,8}{98}=0,6\left(mol\right)\)

Theo PT: \(n_{R_2O_3}=\dfrac{1}{3}n_{H_2SO_4}=0,2\left(mol\right)\)

\(\Rightarrow M_{R_2O_3}=\dfrac{32}{0,2}=160=2M_R+16.3\Rightarrow M_R=56\left(g/mol\right)\)

→ R là Fe.

b, Ta có: m _{dd sau pư} = m_Fe2O3 + m _{dd H2SO4} = 32 + 168 = 200 (g)

Ta có: \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)

Theo PT: \(n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,2\left(mol\right)\)

\(\Rightarrow C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,2.400}{200}.100\%=40\%\)

c, Ta có: \(n_{Fe_2\left(SO_4\right)_3.10H_2O}=n_{Fe_2\left(SO_4\right)_3}=0,2\left(mol\right)\)

\(\Rightarrow m_{Fe_2\left(SO_4\right)_3.10H_2O}=0,2.580=116\left(g\right)\)

Ta có: \(40n_{MgO}+160n_{Fe_2O_3}=3,84\left(1\right)\)

Mà: Oxi chiếm 32,5% về khối lượng.

⇒ m_O = 3,84.32,5% = 1,248 (g) \(\Rightarrow n_O=\dfrac{1,248}{16}=0,078\left(mol\right)\)

\(\Rightarrow n_{MgO}+3n_{Fe_2O_3}=0,078\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{MgO}=0,024\left(mol\right)\\n_{Fe_2O_3}=0,018\left(mol\right)\end{matrix}\right.\)

PT: \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

Theo PT: \(n_{H_2SO_4}=n_{MgO}+3n_{Fe_2O_3}=0,078\left(mol\right)\)

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{0,078}{1}=0,078\left(l\right)=78\left(ml\right)\)

a, PT: \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

\(Fe_3O_4+8HCl\rightarrow FeCl_2+2FeCl_3+4H_2O\)

b, Gọi: \(\left\{{}\begin{matrix}n_{Al_2O_3}=x\left(mol\right)\\n_{Fe_3O_4}=y\left(mol\right)\end{matrix}\right.\) ⇒ 102x + 232y = 7,33 (1)

Ta có: \(m_{ddHCl}=77,41.1,06=82,0546\left(g\right)\Rightarrow m_{HCl}=82,0546.12,9\%=10,585\left(g\right)\)

\(\Rightarrow n_{HCl}=\dfrac{10,585}{36,5}=0,29\left(mol\right)\)

Theo PT: \(n_{HCl}=6n_{Al_2O_3}+8n_{Fe_3O_4}=6x+8y=0,29\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,015\left(mol\right)\\y=0,025\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Al_2O_3}=0,015.102=1,53\left(g\right)\\m_{Fe_3O_4}=0,025.232=5,8\left(g\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al_2O_3}=\dfrac{1,53}{7,33}.100\%\approx20,87\%\\\%m_{Fe_3O_4}\approx79,13\%\end{matrix}\right.\)

c, Ta có: \(n_{Al_2O_3}=0,015.\dfrac{3,665}{7,33}=0,0075\left(mol\right)\)

PT: \(Al_2O_3+2KOH\rightarrow2KAlO_2+H_2O\)

Theo PT: \(n_{KOH}=2n_{Al_2O_3}=0,015\left(mol\right)\)

\(\Rightarrow V_{ddKOH}=\dfrac{0,015}{4}=0,00375\left(l\right)=3,75\left(ml\right)\)

PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

\(2Zn+O_2\underrightarrow{t^o}2ZnO\)

Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}+\dfrac{1}{2}n_{Zn}=\dfrac{3}{4}x+\dfrac{1}{2}y\left(mol\right)\)

\(\Rightarrow\dfrac{3}{4}x+\dfrac{1}{2}y=0,15\)