a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

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

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

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

_____0,2_____0,4__________0,2 (mol)

a, \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

b, \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)

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

Theo PT: \(n_{Fe}=n_{H_2}=0,4\left(mol\right)\)

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

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

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

- Trích mẫu thử.

- Nhỏ từng mẫu thử vào giấy quỳ tím.

+ Quỳ hóa xanh: Ca(OH)₂

+ Quỳ hóa đỏ: H₂SO₄

+ Quỳ không đổi màu: Na₂SO₄

- Dán nhãn.

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

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

\(\Rightarrow m_{Zn}=0,3.65=19,5\left(g\right)\)

c, \(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,1}=6\left(M\right)\)

a, m_Na2CO3 = 0,2.106 = 21,2 (g)

b, n_NaOH = 0,2.1,5 = 0,3 (mol)

⇒ m_NaOH = 0,3.40 = 12 (g)

c, m_HCl = 300.14,6% = 43,8 (g)

Bạn bổ sung đầy đủ đề nhé.

Ta có: 84n_MgCO3 + 100n_CaCO3 = 18,4 (1)

PT: \(MgCO_3+H_2SO_4\rightarrow MgSO_4+CO_2+H_2O\)

\(CaCO_3+H_2SO_4\rightarrow CaSO_4+CO_2+H_2O\)

Theo PT: \(n_{CO_2}=n_{MgCO_3}+n_{CaCO_3}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\left(2\right)\)

Từ (1) và (2) ⇒ n_MgCO3 = n_CaCO3 = 0,1 (mol)

Theo PT: \(\left\{{}\begin{matrix}n_{MgSO_4}=n_{MgCO_3}=0,1\left(mol\right)\\n_{CaSO_4}=n_{CaCO_3}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow C_{M_{MgSO_4}}=C_{M_{CaSO_4}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

a, \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

b, Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,1.65}{10,5}.100\%\approx61,9\%\\\%m_{Cu}\approx38,1\%\end{matrix}\right.\)

c, \(n_{H_2SO_4}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,1}{0,5}=0,2\left(M\right)\)