Zn + 2 HCl -> ZnCl₂ + H₂
a)
Vedem Zn reactioneaza cu HCl in raport de 1:2. Deci la 2 moli de Zn for reactiona 4 moli de HCl. Deci substanta in exces este:
6 moli HCl - 4 moli HCl = 2 moli HCl
b)
Daca trecem substanta in exces prin NaOH pana la saturatie, avem:
NaOH + HCl -> NaCl + H₂O
Vedem ca HCl reactioneaza cu NaOH in raport de 1:1. Deci la 2 moli de HCl trebuie sa avem 2 moli de NaOH.
1 mol NaOH = 40g => 2 moli NaOH = 80 g
80g........................40%
Xg...........................100%
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X = 100 * 80 / 40 = 200g solutie NaOH 40% concentratie
