mKOH có trong dd A là = \(\dfrac{C\%\cdot mdd}{100}=\dfrac{20\cdot70}{100}=14\left(g\right)\)
\(n_K=\dfrac{m}{M}=\dfrac{5,85}{39}=0,15\left(mol\right)\)
PTHH : 2K + 2H2O ---> 2KOH + H2
..............0,15.....0,15.............0,15..............0,15 (mol)
=> \(m_{KOH}=n\cdot M=0,15\cdot56=8,4\left(g\right)\)( tạo ra khi thêm K )
=>tổng mKOH = 14 + 8,4 = 22,4(g)
mddKOH = 70 + 5,85 = 75,85 (g)
=> \(C\%dd_{KOH}=\dfrac{m_{KOH}}{mdd_{KOH}}\cdot100\%=\dfrac{22,4}{75,85}\cdot100\%=29,53\%\)