Chemistry Pre-Lab Question?

Question

A student performing this experiment finds that her crucible and cover weigh 21.234g. After she adds the unknown, the mass goes up to 21.520g. When the sample is converted to a chloride and is dried, she finds that the mass of the crucible, cover, and chloride is 21.533g. She was told that her sample contained Barium, which had a mass percent in the unknown equal to 69.9%.

How many grams of barium chloride are formed?

Any help will be really appreciated and helpful.

Answer 1

21.533 - 21.234 = 0.299 g of product. But if barium is limiting it might be less. So:

21.520 - 21.234 = 0.286 g of unknown

0.286 x (69.9/100) = 0.200 g of barium

Barium chloride is BaCl2 and has a molecular weight of 208.23. Barium has a molecular weight of 137.33.

137.33 / 208.23 x 100 =  65.9% barium.

0.299 x (65.9 / 100) = 0.197 g of barium, which means you can form 0.299 g of barium chloride.