Al2O3 + H2SO4 = Al2(SO4)3 + H2O
Aluminum oxide, Al2O3 would produce aluminum by the following decomposition:2Al2O3 ==> 4Al + 3O2 750.0 g Al2O3 x 1 mole/101.96 g = 7.356 moles Al2O3 Theoretical yield of Al = 7.356 moles Al2O3 x 4 moles Al/2 mole Al2O3 = 14.71 moles Al Theoretical mass of Al = 14.71 moles Al x 26.98 g/mole = 396.9 g Al Percent yield = 256.734 g/396.9 g (x100%) = 64.68% yield (to 4 significant figures)
Ca(OH)2 + H2SO4 >> CaSO4 + 2H2O
H2so4
The balanced chemical equation would be K4FeC6N6 + KMnO4 + H2SO4 = KHSO4 + Fe2SO43 + MnSO4 + HNO3 + CO2 + H2O.
10S + 40 HNO3 → 10 H2SO4 + 40 NO2 + H2OS + 6HNO3 → H2SO4 + 6NO2 + 2H2OS + 4HNO3 -→ SO2 + 4NO2 + 2H2O
Ch3och3
Water and potassium sulfate
It will yield iron sulphate and water.
k2so4+hno2
Al2O3 + 6HBr = 2Albr3 + 3H2O
81.93%
Fe2O3 + H2SO4
MgCl2 + H2SO4 ---> MgSO4 + 2HCl yield magnesium sulfate plus hydrogen chloride(gaz)
Aluminum oxide, Al2O3 would produce aluminum by the following decomposition:2Al2O3 ==> 4Al + 3O2 750.0 g Al2O3 x 1 mole/101.96 g = 7.356 moles Al2O3 Theoretical yield of Al = 7.356 moles Al2O3 x 4 moles Al/2 mole Al2O3 = 14.71 moles Al Theoretical mass of Al = 14.71 moles Al x 26.98 g/mole = 396.9 g Al Percent yield = 256.734 g/396.9 g (x100%) = 64.68% yield (to 4 significant figures)
H2SO4+Hg
Pb + PbO2 + H2SO4 --> PbSO4 + H2O
This equation is Al2O3 + 6 HCl = 2 AlCl3 + 3 H2O.