Koliko je grama 10% sumporne kiseline potrebno za oslobadjanje sircetne kiseline iz 0,2 mola natrijum-acetata?
2CH3COONa+H2SO4--------->2CH3COOH+Na2SO4
n(natrijum-acetata)=0.2 mol
n(natrijum-acetata):n(H2SO4)=2:1
n(H2SO4)=n(natrijum-acetata)*1/2=0.2*1/2=0.1 mol
m(H2SO4)=n(H2SO4)*Mr=0.1*98=9.8 g
w(H2SO4)=m(H2SO4)/mr
mr=m(H2SO4)/w(H2SO4)=9.8/0.1=98 g 10% rastvora
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2CH3COONa+H2SO4--------->2CH3COOH+Na2SO4
n(natrijum-acetata)=0.2 mol
n(natrijum-acetata):n(H2SO4)=2:1
n(H2SO4)=n(natrijum-acetata)*1/2=0.2*1/2=0.1 mol
m(H2SO4)=n(H2SO4)*Mr=0.1*98=9.8 g
w(H2SO4)=m(H2SO4)/mr
mr=m(H2SO4)/w(H2SO4)=9.8/0.1=98 g 10% rastvora