Just make it a regular stoichiometry problem.
CaSO4*2H2O ==> CaSO4 + 2H2O
if you heat 15.00g of the hydrate CaSO4*2H20, and drive off the water, what is the weight of the anhydrous salt remaining?
How do I go about solving this?
3 answers
so then would i find the gfm of the CaSO4?
and get 136.2 as my answer?
and get 136.2 as my answer?
Do you really believe you can start with 15.00 g CaSO4.2H2O, lose the water, and end up with more than 15.00 g? Not likely. It MUST be less than 15.00 grams CaSO4.
CaSO4.2H2O ==> CaSO4 + 2H2O
moles CaSO4.2H2O = 15.00 g x (1 mole CaSO4.2H2O/172.172) = 0.08712 moles.
moles CaSO4 = same.
grams CaSO4 = moles CaSO4 x molar mass CaSO4 = ??
CaSO4.2H2O ==> CaSO4 + 2H2O
moles CaSO4.2H2O = 15.00 g x (1 mole CaSO4.2H2O/172.172) = 0.08712 moles.
moles CaSO4 = same.
grams CaSO4 = moles CaSO4 x molar mass CaSO4 = ??