A properly written formula for the compound, using the subscript fonts available for answers but not for questions, is Al2(SO4)3. This formula shows that there are 2 aluminum atoms, three sulfur atoms, and twelve oxygen atoms in each formula unit, for a total of 342.14 grams per formula unit. 647 of these formula units will have a mass of 647 X 342.14 gramsor 221 kilograms, to the justified number of significant digits.
Aluminum sulfate is formulated by Al2(SO4)3
Well...Since Aluminum is insoluble in water as an element I assume you mean that an 0.12 mol Aluminum ions react with 0.12 mol Sulfate ions to form x mol of Aluminum Sulfate. Balanced equation (Kinda since since the other portions of the compounds are unknown): 2 Al3+(aq) + 3 SO42-(aq) <----->Al2(SO4)3 (aq) 0.12mol*(1mol Al2(SO4)3 (aq)/2mol Al3+)=0.06mol Al2(SO4)3 (aq) 0.12mol*(1molAl2(SO4)3/3mol SO42-) =0.04mol Al2(SO4)3 (aq) *Answer*
3Fe (s) + Al2(SO4)3 (aq) → 3Fe(SO4) (aq) + 2Al (s)
342
Al2(SO4)3
Al2 SO4(wrong) this is how it should be written Al2(SO4)3This how you do it3.5 moles Al2(SO4)3 x 3 mole Sulfur / 1 moles Al2(SO4)3 = put that in ur calculator it will give u the answer
5,7 moles (SO4)3-.
Aluminum sulfate is formulated by Al2(SO4)3
how many moles of sulfur are present in 3.4 moles of Al2(SO4)3
Formula: Al2(SO4)3
How many moles are in 4.70g of NH3?
The formula for the anhydrous form is Al2(SO3)3.
Al2(SO4)3 [anhydrous] = 342,15 Al2(SO4)3 [octadecahydrate] = 648,41
Well...Since Aluminum is insoluble in water as an element I assume you mean that an 0.12 mol Aluminum ions react with 0.12 mol Sulfate ions to form x mol of Aluminum Sulfate. Balanced equation (Kinda since since the other portions of the compounds are unknown): 2 Al3+(aq) + 3 SO42-(aq) <----->Al2(SO4)3 (aq) 0.12mol*(1mol Al2(SO4)3 (aq)/2mol Al3+)=0.06mol Al2(SO4)3 (aq) 0.12mol*(1molAl2(SO4)3/3mol SO42-) =0.04mol Al2(SO4)3 (aq) *Answer*
Aluminum Sulfate
3Fe (s) + Al2(SO4)3 (aq) → 3Fe(SO4) (aq) + 2Al (s)
342