Sửa đề: 9,9161 (l) → 9,916 (l)

a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)

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

Theo PT: n_Fe = n_H2 = 0,4 (mol)

⇒ m_Fe = 0,4.56 = 22,4 (g)

CTHH K₂SO₄ cho biết

- K₂SO₄ do 3 nguyên tố: K, S và O tạo ra.

- K₂SO₄ gồm 2 nguyên tử K, 1 nguyên tử S và 4 nguyên tử O.

- Phân tử khối = 39.2 + 32 + 16.4 = 174 

\(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\)

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

b, \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

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

⇒ V_H2 = 0,3.24,79 = 7,437 (l)

c, \(n_{HCl}=3n_{Al}=0,6\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,2}=3\left(M\right)\)

\(n_{H_2SO_4.2SO_3}=\dfrac{38,7}{258}=0,15\left(mol\right)\)

⇒ BTNT S: n_{H2SO4 (thêm vào)} = 0,15.3 = 0,45 (mol)

\(m_{H_2SO_4\left(30\%\right)}=\dfrac{100.30\%}{98}=30\left(g\right)\)

\(\Rightarrow C\%_{H_2SO_4}=\dfrac{0,45.98+30}{38,7+100}.100\%\approx53,42\%\)

\(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)

a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)

___0,15______0,3__________0,15 (mol)

b, m_HCl = 0,3.36,5 = 10,95 (g)

c, V_H2 = 0,15.24,79 = 3,7185 (l)

Ta có: \(n_{H_2SO_4}=\dfrac{25.19,6\%}{98}=0,05\left(mol\right)\)

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

____1/60________0,05________1/60 (mol)

⇒ m _{dd sau pư} = 1/60.160 + 25 = 83/3 (g)

\(\Rightarrow C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{\dfrac{1}{60}.400}{\dfrac{83}{3}}.100\%\approx24,1\%\)

\(CH_2=CH_2+H_2O\underrightarrow{^{t^o,xt}}C_2H_5OH\)

\(C_2H_5OH+O_2\underrightarrow{^{t^o,mengiam}}CH_3COOH+H_2O\)

\(2CH_3COOH+Ca\left(OH\right)_2\rightarrow\left(CH_3COO\right)_2Ca+2H_2O\)

1. \(C\%_{NaCl}=\dfrac{10}{10+190}.100\%=5\%\)

2. \(n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\)

\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,1}{0,4}=0,25\left(M\right)\)

3. \(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

____0,3______0,6______0,3___0,3 (mol)

a, \(m_{Mg}=0,3.24=7,2\left(g\right)\)

b, \(C_{M_{HCl}}=\dfrac{0,6}{0,2}=3\left(M\right)\)

c, \(m_{MgCl_2}=0,3.95=28,5\left(g\right)\)

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

b, \(Fe\left(OH\right)_3+3HCl\rightarrow FeCl_3+3H_2O\)

c, \(3Ca\left(OH\right)_2+2FeCl_3\rightarrow2Fe\left(OH\right)_3+3CaCl_2\)