In the balanced double replacement reaction of calcium chloride CaCl2 and aluminum sulfate Al2(SO4)3, how many grams of calcium sulfate can be produced if you start the reaction with 4.19 grams of calcium chloride and the reaction goes to completion?
6CaCl2 + 2Al2(SO4)3 = 6CaSO4 + 4AlCl3
So, now you know that each mole of CaCl2 produces one mole of CaSO4
How many moles in 4.19g of CaCl2?
How many g in that much CaSO4?
6 mols?
so 4.19x6?
1 mole of CaCl2 is about 111 g
(Ca = 40, Cl = 35.5)
So, 4.19g is not 6 moles!
so 4.19/111?
that's better
that answer is also wrong though
what am I missing?
Looking at my equation, I see i could have factored out a 2:
3CaCl2 + Al2(SO4)3 = 3CaSO4 + 2AlCl3
that does not affect the relative amounts, however.
4.19/111 = .0377 moles CaCl2
.0377 moles CaSO4 = .0377*136 = 5.13g
Did you forget to calculate the grams, as they asked? You need the relative quantities in moles, but you have to convert grams of input to moles, then back to grams of output.