Chemistry question?

Question

On heating 5.03 g of hydrated barium chloride, 4.23 g is left behind. What is the name of this hydrate?

Answer 1

1) Subtract the two masses and see how much water was driven off when heated:

5.03 g - 4.23 g = 0.8 g H2O given off

2) Change this mass from grams to moles of H2O:

0.8 g H2O / 18 g H2O in 1 mole = 0.044 mol H2O

3) Change left over mass of BaCl2 to moles BaCl2:

4.23 g BaCl2 / 207 g BaCl2 in 1 mol = 0.021 mol BaCl2

4) Get the ratio of mol H2O to mol BaCl2:

0.044 mol H2O : 0.021 mol BaCl2

5) There is a ratio of 2:1 so two H2O for each BaCl2, therefore the hydrate was named:
Barium chloride di-hydrate

Answer 2

i agree with the answerer above me