Question

A student had a mixture of sand and barium chloride dihydrate (BaCl2x2H2O). A beaker containing the mixture weighed 72.248g. when empty the beaker weighed 69.748g. the mixture was determined to contain 45.26% sand.
(a) What is the % and mass of the hydrate in the mixture?
(b) If the mixture was selectively decomposed by heating, how many grams and moles of water would be lost?

Answer 1

Answer:

a) 54.74% is the percentage and 1.3685 grams is the mass of the hydrate in the mixture.

b) Moles of water produced as 0.01122 moles and mass of water produced is 0.2020 grams.

Explanation:

a) Mass of beaker with mixture = x = 72.248 g

Mass of beaker = y = 69.748 g

Mass of mixture = m = x-y = 72.248 g - 69.748 g = 2.5 g

Percentage of sand in mixture = 45.26%

Percentage of barium chloride hydrate in mixture =100 % - 45.26% = 54.74%

Mass of the barium chloride hydrate in the mixture : 54.74% of m

= $\frac{54.74}{100}\times 2.5 g=1.3685 g$

54.74% is the percentage and 1.3685 grams is the mass of the hydrate in the mixture.

b) $BaCl_2.2H_2O\rightarrow BaCl_2+2H_2O$

Moles of barium chloride dihydrate = $\frac{1.3685 g}{244 g/mol}=0.005609 mol$

1 mole of of barium chloride dihydrate has 2 moles of water.Then 0.005609 moles of barium chloride dihydrate will give:

$2\times 0.005609 mol =0.01122 mol$

Mass of 0.01122 moles of water = 0.00122 mmol 18 g/mol = 0.2020 g

Moles of water produced as 0.01122 moles and mass of water produced is 0.2020 grams.