Bài 2 :
\(n_{NaOH}=\dfrac{m}{M}=\dfrac{1,6}{40}=0,04\left(mol\right)\)
PTHH : 2Na + 2H2O ----> 2NaOH + H2
.............0,04......0,04..........0,04.........0,02
=> \(m_{Na}=n\cdot M=0,04\cdot23=0,92\left(g\right)\)
c) \(m_K=0,92\Rightarrow n_K=\dfrac{m}{M}=\dfrac{0,92}{39}\approx0,0236\)
PTHH : 2K + 2H2O ----> 2KOH + H2
..........0,0236...0,0236........0,0236.....0,0118..(mol)
=>\(V_{H_2}=n\cdot22,4=0,0118\cdot22,4=0,26432\left(l\right)\)