ms1 = 80g, c1= 5,6%
ms2 = ?g, c2%= 7,3%
ms3 = ?g,
stim ca c% = mdx100/ms
=> md1 = ms1xc1/100 = 80x5,6/100 = 4,48g KOH
dupa reactie se obtine KCl care este md.final si apa nou formata la care se va adauga si apa din cele doua solutii ms1 si ms2
4,48g md2 md3 m
KOH + HCl --> KCl + H2O
56 36,5 74,5 18
=> md2 = 4,48x36,5/56 = 2,292g HCl consumat
=> ms2 = md2x100/c2 = 2,292x100/7,3 = 40g sol. HCl
=> md3 = 4,48x74,5/56 = 5,96g KCl format
=> m = 4,48x18/56 = 1,44 g apa nou formata
din S1=> m.apa1= 80-4,48 = 75,52g
din S2=> m.apa2 = 40-2,292 = 37,708g
=> m.apa.final = m.apa1 + m.apa2 + m = 1,44+75,52+37,708 = 114,67 g
=> ms.final = md3 + m.apa.final = 5,96 + 114,67 = 120,63 g
=> c%.final = md3x100/ms.final = 5,96x100/120,63 = 4,94%
