Ta có: \(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)

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

_____0,3_____0,9______0,2____0,45 (mol)

a, m_Al = 0,3.27 = 8,1 (g)

b, \(C_{M_{HCl}}=\dfrac{0,9}{0,15}=6\left(M\right)\)

Ta có: \(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

_____0,1_____0,2______0,1___0,1 (mol)

a, m_Mg = 0,1.24 = 2,4 (g)

b, m_HCl = 0,2.36,5 = 7,3 (g)

c, Cách 1: m_MgCl2 = 0,1.95 = 9,5 (g)

Cách 2: Theo ĐLBT KL, có: m_Mg + m_HCl = m_MgCl2 + m_H2

⇒ m_MgCl2 = 2,4 + 7,3 - 0,1.2 = 9,5 (g)

Ta có: \(n_{Cl_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PT: \(Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O\)

____0,2_______0,4______0,2______0,2 (mol)

\(\Rightarrow V_{NaOH}=\dfrac{0,4}{4}=0,1\left(l\right)\)

\(C_{M_{NaCl}}=C_{M_{NaClO}}=\dfrac{0,2}{0,1}=2\left(M\right)\)

Sửa đề: RO → RO₂

Ta có: \(M_{RO_2}=2,285.28\approx64\left(g/mol\right)\)

⇒ M_R = 64 - 16.2 = 32 (g/mol)

→ R là S.

Vậy: CTPT cần tìm là SO₂

a, Ta có: 27n_Al + 56n_Fe = 27,8 (1)

PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

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

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{17,353}{24,79}=0,7\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,2\left(mol\right)\\n_{Fe}=0,4\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{27,8}.100\%\approx19,42\%\\\%m_{Fe}\approx80,58\%\end{matrix}\right.\)

b, \(n_{H_2SO_4}=n_{H_2}=0,7\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,7}{0,5}=1,4\left(M\right)\)

Bạn tách câu hỏi ra nhé.

a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)

b, \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Theo PT: \(n_{Mg}=n_{H_2}=0,5\left(mol\right)\Rightarrow m_{Mg}=0,5.24=12\left(g\right)\)

\(\Rightarrow m_{Cu}=20-12=8\left(g\right)\)

c, \(n_{HCl}=2n_{H_2}=1\left(mol\right)\Rightarrow V_{HCl}=\dfrac{1}{1}=1\left(l\right)\)

Đáp án: C

Base không tan: Mg(OH)₂, Cu(OH)₂, Fe(OH)₂, Fe(OH)₃.

Bạn tách câu hỏi ra nhé.

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

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

\(n_{ZnCl_2}=n_{H_2}=0,2\left(mol\right)\Rightarrow m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)