K2O + H2O → 2KOH (1)
a) \(n_{K_2O}=\dfrac{47}{94}=0,5\left(mol\right)\)
Theo PT1: \(n_{KOH}=2n_{K_2O}=2\times0,5=1\left(mol\right)\)
\(\Rightarrow C_{M_{KOH}}=\dfrac{1}{0,5}=2\left(M\right)\)
c) 2KOH + H2SO4 → K2SO4 + 2H2O
Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{KOH}=\dfrac{1}{2}\times1=0,5\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,5\times98=49\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{49}{20\%}=245\left(g\right)\)
\(\Rightarrow V_{ddH_2SO_4}=\dfrac{245}{1,14}=214,912\left(l\right)\)