Koliko je grama 10% sumporne kiseline potrebno za oslobadjanje sirćetne kiseline iz 0,2mola natrijum-acetata? Jednačina mi treba...
2CH3COONa+H2SO4--------->2CH3COOH+Na2SO4
n(natrijum-acetata):n(H2SO4)=2:1
n(H2SO4)=n(natrijum-acetata)*1/2=0.2*1/2=0.1mol
m(H2SO4)=n(H2SO4)*Mr=0.1*98=9.8g
w(H2SO4)=m(H2SO4)/mr
mr=m(H2SO4)/w(H2SO4)=9.8/0.1=98g 10% H2SO4
powered by Drupal
SIRCETNA KISELINA
2CH3COONa+H2SO4--------->2CH3COOH+Na2SO4
n(natrijum-acetata):n(H2SO4)=2:1
n(H2SO4)=n(natrijum-acetata)*1/2=0.2*1/2=0.1mol
m(H2SO4)=n(H2SO4)*Mr=0.1*98=9.8g
w(H2SO4)=m(H2SO4)/mr
mr=m(H2SO4)/w(H2SO4)=9.8/0.1=98g 10% H2SO4