A 0.940g sample of barium chloride dehydrate is heated and 0.800g of anhydrous residue remains after cooling.

Question

A. How many moles of anhydrous barium chloride were present in the sample?
B. How many moles of water were present in the sample?

Answer 1

A.
(0.800g BaCl2) / (208.233g BaCl2/mol) =
0.00384 mol BaCl2

B.
(0.940 g - 0.800 g) /
(18.01532 g H2O/mol) = 0.00777 mol H2O

Answer 2

I don't answer the answer that's why I posted the question.