Ba(ClO3)2 + H2SO4 = 2 HClO3 + BaSO4Izracunajte koliko je grama barijum-hlorata potrebno za dobijanje 200g 40% HLORATNE KISELINE.
Ba(ClO3)2+H2SO4---------->2HClO3+BaSO4
w=ms/mr
ms=w*mr=0.40*200=80 g hloratne kiseline
n(HClO3)=m/Mr=80/84.5=0.95 mol
n(HClO3):n(Ba(ClO3)2)=2:1
n(Ba(ClO3)2)=n(HClO3)*1/2=0.95*1/2=0.475 mol
m(Ba(ClO3)2)=n(Ba(ClO3)2)*Mr=0.475*304=144.4 g
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Ba(ClO3)2+H2SO4---------->2HC
Ba(ClO3)2+H2SO4---------->2HClO3+BaSO4
w=ms/mr
ms=w*mr=0.40*200=80 g hloratne kiseline
n(HClO3)=m/Mr=80/84.5=0.95 mol
n(HClO3):n(Ba(ClO3)2)=2:1
n(Ba(ClO3)2)=n(HClO3)*1/2=0.95*1/2=0.475 mol
m(Ba(ClO3)2)=n(Ba(ClO3)2)*Mr=0.475*304=144.4 g