a, Chất rắn là Ag.

⇒ m_Ag = 10,8 (g)

PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

Ta có: \(n_{H_2}=\dfrac{14,874}{24,79}=0,6\left(mol\right)\)

Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,4\left(mol\right)\)

⇒ m_Al = 0,4.27 = 10,8 (g)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{10,8}{10,8+10,8}.100\%=50\%\\\%m_{Ag}=50\%\end{matrix}\right.\)

b, Theo PT: \(n_{HCl}=2n_{H_2}=1,2\left(mol\right)\)

\(\Rightarrow V=\dfrac{1,2}{1,5}=0,8\left(l\right)=800\left(ml\right)\)

\(n_{AlCl_3}=\dfrac{2}{3}n_{H_2}=0,4\left(mol\right)\Rightarrow C_{M_{AlCl_3}}=\dfrac{0,4}{0,8}=0,5\left(M\right)\)

a, \(2Al+3Cl_2\underrightarrow{t^o}2AlCl_3\)

b, \(n_{Al}=\dfrac{12,15}{27}=0,45\left(mol\right)\)

\(n_{Cl_2}=\dfrac{14,874}{24,79}=0,6\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,45}{2}>\dfrac{0,6}{3}\), ta được Al dư.

Theo PT: \(n_{Al\left(pư\right)}=n_{AlCl_3}=\dfrac{2}{3}n_{Cl_2}=0,4\left(mol\right)\)

⇒ n_{Al (dư)} = 0,45 - 0,4 = 0,05 (mol)

⇒ m_{Al (dư)} = 0,05.27 = 1,35 (g)

b, m_AlCl3 = 0,4.133,5 = 53,4 (g)

Hỗn hợp khí thu được gồm: CO và CO₂

Ta có: \(\left\{{}\begin{matrix}n_{CO}+n_{CO_2}=\dfrac{4,958}{24,79}=0,2\\28n_{CO}+44n_{CO_2}=19.2.0,2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{CO}=0,075\left(mol\right)\\n_{CO_2}=0,125\left(mol\right)\end{matrix}\right.\)

PT: \(C+\dfrac{1}{2}O_2\underrightarrow{t^o}CO\)

\(C+O_2\underrightarrow{t^o}CO_2\)

Theo PT: \(n_C=n_{CO}+n_{CO_2}=0,2\left(mol\right)\)

⇒ m_C = 0,2.12 = 2,4 (g)

Ta có: m _muối = m_KL + m_Cl

⇒ 72,375 = 24,45 + 35,5n_Cl

⇒ n_Cl = 1,35 (mol)

BTNT Cl: n_HCl = n_{Cl (trong muối)} = 1,35 (mol)

⇒ m_HCl = 1,35.36,5 = 49,275 (g)

Ta có: 12n_C + 32n_S = 10 (1)

PT: \(C+O_2\underrightarrow{t^o}CO_2\)

\(S+O_2\underrightarrow{t^o}SO_2\)

Theo PT: \(n_{O_2}=n_C+n_S=\dfrac{12,395}{24,79}=0,5\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_C=0,3\left(mol\right)\\n_S=0,2\left(mol\right)\end{matrix}\right.\)

⇒ m_C = 0,3.12 = 3,6 (g)

m_S = 0,2.32 = 6,4 (g)

Theo PT: \(n_{CO_2}+n_{SO_2}=n_{O_2}=0,5\left(mol\right)\)

\(\Rightarrow V_{hhk}=0,5.24,79=12,395\left(l\right)\)

a, \(n_{Mg}=\dfrac{5}{24}\left(mol\right)\)

\(n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

Xét tỉ lệ: \(\dfrac{\dfrac{5}{24}}{1}>\dfrac{0,4}{2}\), ta được Mg dư.

Theo PT: \(n_{Mg\left(pư\right)}=\dfrac{1}{2}n_{HCl}=0,2\left(mol\right)\)

⇒ m_{Mg (pư)} = 0,2.24 = 4,8 (g)

⇒ m_{Mg (dư)} = 5 - 4,8 = 0,2 (mol)

b, \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,2\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)

a, Ta có: 27n_Al + 56n_Fe = 11 (1)

PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

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

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{9,916}{24,79}=0,4\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,2\left(mol\right)\\n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{11}.100\%\approx49,09\%\\\%m_{Fe}\approx50,91\%\end{matrix}\right.\)

b, Theo PT: \(n_{HCl}=2n_{H_2}=0,8\left(mol\right)\)

\(\Rightarrow V_{HCl}=\dfrac{0,8}{2}=0,4\left(l\right)\)

a, Ta có: 24n_Mg + 27n_Al = 12,9 (1)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

Theo PT: \(\left\{{}\begin{matrix}n_{MgCl_2}=n_{Mg}\\n_{AlCl_3}=n_{Al}\end{matrix}\right.\) ⇒ 95n_Mg + 133,5n_Al = 59,05 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,2\left(mol\right)\\n_{Al}=0,3\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,2.24}{12,9}.100\%\approx37,21\%\\\%m_{Al}\approx62,79\%\end{matrix}\right.\)

b, Theo PT: \(n_{HCl}=2n_{Mg}+3n_{Al}=1,3\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{1,3.36,5}{14,6\%}=325\left(g\right)\)

c, Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,65\left(mol\right)\)

Ta có: m _{dd sau pư} = 12,9 + 325 - 0,65.2 = 336,6 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,2.95}{336,6}.100\%\approx5,6\%\\C\%_{AlCl_3}=\dfrac{0,3.133,5}{336,6}.100\%\approx11,9\%\end{matrix}\right.\)

e, \(S+O_2\underrightarrow{t^o}SO_2\)

\(2SO_2+O_2\underrightarrow{^{t^o,V_2O_5}}2SO_3\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

\(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

\(Na_2SO_4+BaCl_2\rightarrow2NaCl+BaSO_4\)

a, X là Aluminium oxide (Al₂O₃)

b, Ta có: \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)

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

\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,15\left(mol\right)\)

Ta có: n_H2SO4 = 0,4.1,5 = 0,6 (mol)

Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,6}{3}\), ta được H₂SO₄ dư.

Theo PT: \(\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=n_{Al_2O_3}=0,15\left(mol\right)\\n_{H_2SO_4\left(pư\right)}=3n_{Al_2O_3}=0,45\left(mol\right)\end{matrix}\right.\)

⇒ n_{H2SO4 (dư)} = 0,6 - 0,45 = 0,15 (mol)

\(\Rightarrow C_{M_{H_2SO_4}}=C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,15}{0,4}=0,375\left(M\right)\)