NaOH : KOH = 1x : 2x , unde x = multiplu comun
1x = 3 => x = 3
=> amestecul contine 3 moli NaOH + 6 moli KOH
M.NaOH = 23+16+1 = 40 g/mol
M.KOH = 39+16+1 = 56 g/mol
3 moli *40g NaOH = 120 g
6 moli * 56g = 336g
=> m.amestec = 120+336 = 456 g
la 40+56 g NaOH+KOH ........................... 2x16g O
456g amestec NaOH+KOH ................... m = 152 g O
