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

b, Ta có: \(n_{Al}=\dfrac{13,5}{27}=0,5\left(mol\right)\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,75\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,75.24,79=18,5925\left(l\right)\)

c, \(n_{HCl}=3n_{Al}=1,5\left(mol\right)\)

\(\Rightarrow V_{HCl}=\dfrac{1,5}{1}=1,5\left(l\right)=1500\left(ml\right)\)

d, \(n_{AlCl_3}=n_{Al}=0,5\left(mol\right)\)

\(\Rightarrow C_{M_{AlCl_3}}=\dfrac{0,5}{1,5}=\dfrac{1}{3}\left(M\right)\)

Ta có: \(n_{Al}=\dfrac{21,6}{27}=0,8\left(mol\right)\)

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

_____0,8_____2,4_____________1,2 (mol)

a, m_HCl = 2,4.36,5 = 87,6 (g)

b, V_H2 = 1,2.24,79 = 29,748 (l)

Ta có: n_Ca(OH)2 = 0,1.2 = 0,2 (mol)

PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

_____0,2________0,2_______0,2 (mol)

a, V = 0,2.24,79 = 4,958 (l)

b, m_CaCO3 = 0,2.100 = 20 (g) 

Các thể tích đo trong cùng điều kiện nhiệt độ, áp suất.

⇒ Tỉ lệ thể tích cũng là tỉ lệ số mol.

Ta có: V_{O (trong A)} + 2V_O2 = 2V_CO2 + V_H2O

⇒ V_{O (trong A)} = 100

Gọi CTPT của A là C_xH_yO_z.

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{V_{CO_2}}{V_A}=\dfrac{200}{100}=2\\y=\dfrac{2V_{H_2O}}{V_A}=\dfrac{2.200}{100}=4\\z=\dfrac{V_O}{V_A}=\dfrac{100}{100}=1\end{matrix}\right.\)

Vậy: CTPT của A là C₂H₄O.

Ta có: 392n_Cr2(SO4)3 + 152n_FeSO4 = 67,2 (1)

PT: \(2Cr+3H_2SO_4\rightarrow Cr_2\left(SO_4\right)_3+3H_2\)

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

Theo PT: \(n_{H_2SO_4}=3n_{Cr_2\left(SO_4\right)_3}+n_{FeSO_4}=0,32.1,5=0,48\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Cr_2\left(SO_4\right)_3}=0,09\left(mol\right)\\n_{FeSO_4}=0,21\left(mol\right)\end{matrix}\right.\)

Theo PT: n_Cr = 2n_Cr2(SO4)3 = 0,18 (mol)

n_Fe = n_FeSO4 = 0,21 (mol)

⇒ n_Cr:n_Fe = 0,18:0,21 = 6:7

\(n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

Theo PT: \(n_{HCl}=2n_{Mg}=0,5\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,5}{0,1}=5\left(M\right)\)

Ta có: \(n_{CO_2}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)

\(n_{Ca\left(OH\right)_2}=0,15.1=0,15\left(mol\right)\)

\(\Rightarrow\dfrac{n_{CO_2}}{n_{Ca\left(OH\right)_2}}=\dfrac{0,2}{0,15}=1,33\)

→ Pư tạo 2 muối: Ca(HCO₃)₂ và CaCO₃.

PT: \(2CO_2+Ca\left(OH\right)_2\rightarrow Ca\left(HCO_3\right)_2\)

\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=2n_{Ca\left(HCO_3\right)_2}+n_{CaCO_3}=0,2\\n_{Ca\left(OH\right)_2}=n_{Ca\left(HCO_3\right)_2}+n_{CaCO_3}=0,15\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Ca\left(HCO_3\right)_2}=0,05\left(mol\right)\\n_{CaCO_3}=0,1\left(mol\right)\end{matrix}\right.\)

m _{dd sau pư} = m_CO2 + m _ddCa(OH)2 - m_CaCO3 = 0,2.44 + 150.1,02 - 0,1.100 = 151,8 (g)

\(\Rightarrow C\%_{Ca\left(HCO_3\right)_2}=\dfrac{0,05.162}{151,8}.100\%\approx5,3\%\)

\(n_{SO_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)

\(n_{KOH}=0,4.1=0,4\left(mol\right)\)

Ta có: \(\dfrac{n_{KOH}}{n_{SO_2}}=\dfrac{0,4}{0,3}=1,33\)

→ Pư tạo 2 muối: KHSO₃ và K₂SO₃

PT: \(SO_2+KOH\rightarrow KHSO_3\)

\(SO_2+2KOH\rightarrow K_2SO_3+H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{SO_2}=n_{KHSO_3}+n_{K_2SO_3}=0,3\\n_{KOH}=n_{KHSO_3}+2n_{K_2SO_3}=0,4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{KHSO_3}=0,2\left(mol\right)\\n_{K_2SO_3}=0,1\left(mol\right)\end{matrix}\right.\)

m _{dd sau pư} = m_SO2 + m _ddKOH = 0,3.64 + 400.0,8 = 339,2 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{KHSO_3}=\dfrac{0,2.120}{339,2}.100\%\approx7,1\%\\C\%_{K_2SO_3}=\dfrac{0,1.158}{339,2}.100\%\approx4,7\%\end{matrix}\right.\)

a, \(n_{H_2}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)

PT: \(M+2HCl\rightarrow MCl_2+H_2\)

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

\(\Rightarrow M_M=\dfrac{13}{0,2}=65\left(g/mol\right)\)

b, \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\Rightarrow V_{HCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)

c, \(n_{ZnCl_2}=n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

Câu 1:

a, Bạn tự xem lại lí thuyết nhé.

b, Acid: H₂SO₄, HCl, H₂S.

Câu 3:

Ta có: \(n_{H_2}=\dfrac{9,916}{24,79}=0,4\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

a, \(n_{Fe}=n_{H_2}=0,4\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,4.56=22,4\left(g\right)\)

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

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,8}{0,2}=4\left(M\right)\)