\(n_K=0,1\left(mol\right)\)
\(2K\left(0,1\right)+2H_2O\rightarrow2KOH\left(0,1\right)+H_2\left(0,05\right)\)
Dung dịch A là KOH
Khí B là H2
\(\Rightarrow n_{H_2}=0,05\left(mol\right)\)
\(\Rightarrow V_{H_2}\left(đktc\right)=0,05.22,4=1,12\left(l\right)\)
\(n_{KOH}=0,1\left(mol\right)\)
\(100gH_2O=100mlH_2O\)
\(\Rightarrow C_{M_{KOH}}=\dfrac{0,1}{0,1}=1\left(M\right)\)
\(m_{ddA}=m_K+m_{H_2O}-m_{H_2}=2,3+100-0,05.2=102,2\left(g\right)\)
\(\Rightarrow C\%_{KOH}=\dfrac{0,1.56.100}{102,2}=5,48\%\)