HCl + NaOH ----> NaCl + H2O
36,5g 40g 58,5g 18g
x 60g y z
mdHCl=200*36,5/100=73 g
ms NaOH= V*p=299,998g =300g
md NaOH=60g
HCl este in exces !!!!!!
din ec reactiei ==> x=54,75 g HCl ; y=87,75g NaCl; z= 27g H2O
mHCl exces= 73- 54,75=18,25 g
in solutia finala exista NaCl si HCl
masa solutiei finale este :
ms fin=86gH2O+14gNaCl+200gsol HCl+300gsol NaOH= 600g
sau pt. verificare , se mai poate calcula astfel !!!!!!
ms fin= 86 g H2O+ 14 g NaCl+ 87,75 g NaCl(din r.) + 27 g H2O(din r.)+ 18,25 g HCl ex.+ 127g H2O(sol HCl)+240gH2O(solNaOH) =600 g
c% NaCl=(14+87,75)*100/600=......% NaCl
c% HCl=18,25 *188/600=...... % HCl