Answer:
BaCl2.2H2O
Explanation:
Data obtained from the question include:
Mass of barium chloride hydrate (BaCl2.nH2O) = 7.62g
Mass of anhydrous barium chloride (BaCl2) = 6.48g
Next, we shall determine the mass of water in the barium chloride hydrate (BaCl2.nH2O). This can be obtained as follow:
Mass of water (H2O) = Mass of barium chloride hydrate (BaCl2.nH2O) – Mass of anhydrous barium chloride (BaCl2)
Mass of water (H2O) = 7.62 – 6.48
Mass of water (H2O) = 1.14g
Next, we shall write the balanced equation for the reaction. This is given below:
BaCl2.nH2O —> BaCl2 + nH2O
Next, we shall determine the masses of BaCl2.nH2O that decomposed and the mass H2O produced from the balanced equation.
Molar mass of BaCl2.nH2O = 137 + (35.5x2) + n[(2x1) + 16]
= 137 + 71 + n[2 + 16]
= (208 + 18n) g/mol
Mass of BaCl2.nH2O from the balanced equation = 1 x (208 + 18n) = (208 + 18n) g
Molar mass of H2O = (2x1) + 16 = 18 g/mol
Mass of H2O from the balanced equation = n x 18 = 18n g
Summary:
From the balanced equation above,
(208 + 18n) g of BaCl2.nH2O decomposed to produce 18n g of H2O.
Finally, we shall determine the formula for the hydrated compound as follow:
From the balanced equation above,
(208 + 18n) g of BaCl2.nH2O decomposed to produce 18n g of H2O.
But 7.62g of BaCl2.nH2O decomposed to produce 1.14g of H2O i.e
18n/(208 + 18n) = 1.14/7.62
Cross multiply
18n x 7.62 = 1.14(208 + 18n)
137.16n = 237.12 + 20.52n
Collect like terms
137.16n – 20.52n = 237.12
116.64n = 237.12
Divide both side by the coefficient of n i.e 116.64
n = 237.12/116.64
n = 2
Therefore the formula for the hydrate, BaCl2.nH2O is BaCl2.xH2O.
