Question

Salutare va rog ajutati-ma.
m s1=1,6kg NaOH
c1=60% NaOH
m s2=36,5g HCl
c2=36,5% HCl
Sa se calculeze:
a) masa in exces
b) concentratie solutie NaCl
c) masa apa.

Answer 1

HCl+NaOH -> NaCl+H2O

c=100md/ms=> md1=cms1/100=60*1600/100=960 g NaOH

md2=cms2/100=36.5*36.5/100=13,325 g HCl

MHCl=36.5 g/mol
MNaOH= 16+1+23=40 g/mol
MNaCl=35.5+23=58,5 g/mol

a) Ca sa aflam excesul, consideram masa de NaOH necunoscuta.

Punem pe reactie => x/40= 13,325/36.5=> x=0.365*40=14,6 g pentru ca reactia sa fie stoechiometrica => avem un exces de NaOH de 960-14.6=945.4 g

b) 14.6/40=y/58.5 => y=21,3525 g NaCl (md)

c) mH2O/18=14.6/40=0,365 => MH2O=18*0,365=6,57 g apa

Continuarea pentru b) ms=md+mapa=21,3525+6,57=27,9225 g 

c=100md/ms=100*2135,25/27,9225=76,47 %