Question

300 grame de solutie soda caustică de c=20% se neutralizeaza cu solutia de acid clorhidric de c=36,5%. Calculati:
a) masa solutiei de acid clorhidric
b) nr. moli de sare ce rezulta
c) concentratia procentuala a solutiei finale? (c=..%)

Answer 1

NaOH+HCl->NaCl +H²O CNaOH=20% MsNaOH=300 Md=300•20/100 Md=60 MNaOH=16+1+23=40 u.a.m. MHCl= 1+35,5=36,5 u.a.m. MdHCl=30•36,5/40 Md=23,57. MsHCl =23,57•100/36,5 Ms=64,57. MNaCl=23+35,5=58,5 MdNaCl= 30•58,5/40 Md=43,87 N(nr.moli) =43,87/58,5 n=0,74

Answer 2

NaOH+ HCl---->NaCl+H2O

40g     36,5g      1mol   18g

60g       x                y        z            (md NaOH=60g)

-in functie de mdNaOH din ecuatia reactiei se calculeaza :

mHCl (md) , mNaCl si mH2O

x=......g(md) HCl  ;  msHCl= md*100/c=......  ;  mH2O(din solutie)= msHCl-mdHCl

mNaCl=n*M NaCl= 1,5*58,5=87,75g (md)

m s final= m NaCl+ mH2O(din reactie)+ mH2O( din sol. NaOH)+ mH2O(din sol. HCl)=....

c%= mdNaCl*100/ ms final=......