md1=170×2÷100=3,4g AgNO₃
md2=30×14,6÷100=4,38g HCl
MAgNO₃=108+14+48=170g/mol
MHCl=1+35,5=36,5g/mol
reactia 1:1 => 1 mol de AgNO₃:1 mol de HNO₃
170g azotat..................36,5g acid
3,4g azotat..................... Xg acid => X=3,4×36,5÷170=0,73g acid reactionati => 3,65g acid nereactionat
170g 143,5g 63g
AgNO₃ + HCl -> AgCl + HNO₃
3,4g y=2,87g z=1,26g
Insa raman si 3,65g nereactionati, iar apa din solutiile initiale, este si ea prezenta.
m apa din solutia de azotat=170-3,4=166,6g apa
m apa din solutia de acid=30-4,38=25,62g apa
masa de apa totala=192,22g apa.
masa de substante din solutia finala=mdf=2,87+1,26+3,65=7,78g
m solutiei finale=msf=mdf+mapa=7,78+192,22=200g
cf = [tex] \frac{mdf*100}{msf} = \frac{7,78*100}{200}[/tex] = 3,89%
asta se cere defapt!!
Insa pot calcula si compozitia de acizi din solutie.
200g................100%
3,65g.................p => p=1,825% acid clorhidric
1,26g.................t => t=0,63% acid azotic
