\(n_{K_2O}=\dfrac{2,35}{94}=0,025\left(mol\right)\\ K_2O+H_2O\rightarrow2KOH\\ n_{KOH}=0,025.2=0,05\left(mol\right)\\ V_{ddKOH}=V_{ddH_2O}=400\left(ml\right)=0,4\left(l\right)\\ C_{MddKOH}=\dfrac{0,05}{0,4}=0,125\left(M\right)\)
400ml = 0,4l
\(n_{K2O}=\dfrac{2,35}{94}=0,025\left(mol\right)\)
Pt : \(K_2O+H_2O\rightarrow2KOH|\)
1 1 2
0,025 0,05
\(n_{KOH}=\dfrac{0,025.2}{1}=0,05\left(mol\right)\)
\(C_{M_{ddKOH}}=\dfrac{0,05}{0,4}=0,125\left(M\right)\)
Chúc bạn học tốt
K2O+H2O->2KOH
0,025-----------0,05 mol
n K2O=2,35\94=0,025 mol
=>CmKOH=0,05\0,4=0,125M
\(n_{K_2O}=\dfrac{2,35}{94}=0,025\left(mol\right)\)
\(\Rightarrow C_{M_{ddK_2O}}=\dfrac{0,025}{0,4}=0,0625M\)
Đề phải là 4,7 g K2O mới đúng đc
