presupunem masa amestecului 100g
M.molara NaOH = 40 g/mol
M.molara KOH = 56 g/mol
x=nr.moli KOH
y=nr. moli NaOH
56x+40y=100
=> 40y g NaOH contin 9,2 g Na
40g NaOH........23g Na
z g NaOH.........9,2g Na
z=16 g NaOH sau 0,4 moli NaOH (y)
mKOH=100-16=84 g KOH 1,5 moli KOH (x)
x/y=1,5/0,4=3,75