Astudent 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?

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

=

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

b)

Moles of barium chloride dihydrate =

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

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.

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :