Bài 3: Gọi oxit KL cần tìm là A2O3.
Ta có: \(n_{HCl}=\dfrac{60.36,5\%}{36,5}=0,6\left(mol\right)\)
PT: \(A_2O_3+6HCl\rightarrow2ACl_3+3H_2O\)
Theo PT: \(n_{A_2O_3}=\dfrac{1}{6}n_{HCl}=0,1\left(mol\right)\)
\(\Rightarrow M_{A_2O_3}=\dfrac{10,2}{0,1}=102\left(g/mol\right)\)
⇒ 2MA + 16.3 = 102 ⇒ MA = 27 (g/mol)
→ A là Al. Oxit KL cần tìm là Al2O3.
Bài 4:
Ta có: \(n_{K_2O}=\dfrac{18,8}{94}=0,2\left(mol\right)\)
PT: \(K_2O+H_2O\rightarrow2KOH\)
____0,2____________0,4 (mol)
⇒ n KOH (trong 50 ml) = 0,4/10 = 0,04 (mol)
PT: \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
______0,04____0,02 (mol)
\(\Rightarrow m_{H_2SO_4}=0,02.98=1,96\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{1,96}{9,8\%}=20\left(g\right)\)
