a) K2O + H2O ➜ 2KOH
b) \(n_{K_2O}=\frac{94}{94}=1\left(mol\right)\)
Theo PT: \(n_{KOH}=2n_{K_2O}=2\times1=2\left(mol\right)\)
\(\Rightarrow m_{KOH}=2\times56=112\left(g\right)\)
\(C\%_{KOH}=\frac{112}{200}\times100\%=56\%\)
\(V_{ddKOH}=\frac{2}{1}=2\left(l\right)\)
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