I have to assume you mean 2.684 grams of Al2(SO4)3 and water per cc. One of those grams is water so we have 1.684 grams of Al2(SO4)3 in every cc
Al2(s04)3
How many grams per mol?
2 Al = 2 * 27 = 54
3 S = 3 * 32 =
12 O = 12 * 16 = 192
so
Al2 (SO3)4 = 278 grams/mol
If we have 1.684 grams of Al2(SO4)3 per cc, and we know there are 1000 cc per liter then we have 1684 g/liter
1684/278 = 6.06 mols of Al2(SO4)3 in this liter. So I agree with about 6 M
Al2(SO4)3 + 3BaCl2 -> 3BaSO4 + 2 AlCl3
in other words to balance the (SO4)3 we need 3 BaSO4 for every Al2(SO4)3
so 3*9.65 = 29 mols
Al2(s04)3 solution of a molal concentration is present in 1 litre solution of 2.684g/cc. How many moles of bas04 would be precipitated on adding bacl2 in excess. The answer 6M
I want some tutors to explain for me step by step clearly. I do not still understand
Thank a lot all tutors
2 answers
sorry
3 * 6 = 18 mols
3 * 6 = 18 mols
1 answer