Koliko cm3,30% sumporne kiseline,gustine 1,35g\cm3 je potrebno za potpunu neutralizaciju 10g aluminijum hidroksida?
2Al(OH)3+3H2SO4------>Al2(SO4)3+6H2O
n(Al(OH)3)=m/Mr=10/78=0.128mol
n(Al(OH)3):n(H2SO4)=2:3
n(H2SO4)=3n(Al(OH)3)/2=3*0.128/2=0.192mol
m(H2SO4)=n*Mr=0.192*98=18.82g
m(rastvora)=m(H2SO4)/w=18.82/0.30=62.72g
V(H2SO4)=m/q=62.72/1.35=46.46cm3
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Zapremina sump.kiseline
2Al(OH)3+3H2SO4------>Al2(SO4)3+6H2O
n(Al(OH)3)=m/Mr=10/78=0.128mol
n(Al(OH)3):n(H2SO4)=2:3
n(H2SO4)=3n(Al(OH)3)/2=3*0.128/2=0.192mol
m(H2SO4)=n*Mr=0.192*98=18.82g
m(rastvora)=m(H2SO4)/w=18.82/0.30=62.72g
V(H2SO4)=m/q=62.72/1.35=46.46cm3