-masa si moli de NaOH din solutie
c= mdx100/ms---> md= 10x150/100g= 15g
ms= md+m,apa----> m,apa= 150-15=<135g
n= m/M= 15/40 mol= 0,375mol NaOH
-masa si moli HCL din solutia data
md= 36,5x150/100= 54,75g
m,apa= 150-54,75=>95,25g
n= 54,75./36,5-= 1,5mol HCl
-produsii care rezulta din reactie se calculeaza in functie de valoarea cea mai mica
1mol.....1mol........1mol....1mol
HCl + NaOH---> NaCl + H2O
x.........0,375mol......x........x
x= 0,375 mol
m=nxM---> m,NaCl= 0,375x58,5 g=21,94g
m,H20= 0,375X18 g= 6,75g
m,HCl exces= (1.5- 0,375)x36,5= 41 g
solutia este alcatuita din 21,94g NaCl, 41g HcL SI (6,75+135+95,25)g APA
daca calculez c% in NaCl'
md=21,94g
ms= 21,94+41+ 237=300g
c= mdx100/ms= 21,94x100/300=7,3%
