1.
ms1 = 170g , c1% = 60%
ms2 = ? , c2% = 25%
stim ca c% = mdx100/ms
si ca ms = md + m.apa (a)
din sol initiala avem
ca md = ms1xc1%/100
= 170x60/100 = 102 g
25 = 102x100/(170+a)
=> 25(170+a) = 10200
=> a = 238 g apa adaugata
2.
ms1 = 30 kg , c1% = 12%
ms2 = ? c2% = 15%
stim ca c% = mdx100/ms
si ca ms2 = ms1 + m.zahar.adaugat (a)
md1 = ms1xc1%/100
= 30x12/100 = 3,6 kg
15 = (3,6+a)100/(30+a)
=> 15(30+a) = 360+100a
=> a = 1,06 kg zahar adaugat
facem proba:
md.final = 3,6+1,06 = 4,66 kg
ms.final = 30+1,06 = 31,06 kg
=> c.final = 4,66x100/31,06 = 15%
3.
verificam daca reactia a decurs la echivalenta, adica sa nu mai avem exces dintr-un rectant
md.HCl = msxc%/100
= 200x36,5/100 = 73 g HCl
md.NaOH msxc%/100
= 400x20/100 = 80 g NaOH
80 g 73 g
NaOH + HCl --> NaCl + H2O
40 36,5
pt. NaOH = 80/40 = 2
pt HCl = 73/36,5 = 2
=> reactia a decurs la echivalenta si deci nu avem exces de reactant
80 g md g m g
NaOH + HCl --> NaCl + H2O
40 58,5 18
=> md = 80x58,5/40 = 117 g sare NaCl
m = 80x18/40 = 36 g apa
dupa reactie masa de apa din solutiile de HCl si NaOH s-a acimilat la care s-a mai adaugat si m rezultata din reactie
m.apa.NaOH = 400-80 = 320 g
m.apa.HCl = 200 - 73 = 127 g
=> ms.finala = m.apa.HCl + m.apa.NaOH + m + md.sare
= 320 + 127 + 36 + 117 = 600 g sol.
=> c%.finala = 117x100/600 = 19,5%
