A 1.80 g sample of barium chloride hydrate, BaCl2 ⋅ xH2O is treated with excess sulfuric acid, forming a BaSO4 precipitate which has a mass of 1.72 g. Calculate the value of x.

X = 1553 benacne vamsy ue etanbe

The value of x is 2

Explanation:

BaCl2. xH2O react with H2SO4 according to the equation:

BaCl2. xH2O + H2SO4 - > BaSO4 + 2HCl + xH2O

Data obtained from the question include:

Mass of hydrated salt (BaCl2. xH2O) = 1.80g

Mass of BaSO4 = 1.72g

Now, let us calculate the number of mole of BaSO4 present in 1.72g of BaSO4. This is illustrated below:

Molar Mass of BaSO4 = 137 + 32 + (16x4) = 137 + 32 + 64 = 233g/mol

Mass of BaSO4 = 1.72g

Number of mole of BaSO4 = ?

Number of mole = Mass / Molar Mass

Number of mole BaSO4 = 1.72/233

Number of mole BaSO4 = 0.0074 mole.

From the equation above,

1 mole of BaCl2. xH2O produced 1 mole of BaSO4.

Therefore, 0.0074 mole of BaCl2. xH2O will also produce 0.0074 mole of BaSO4.

Now, with this amount (i. e 0.0074 mole) of BaCl2. xH2O, we can obtain the value of x as follow:

Molar Mass of BaCl2. xH2O = 137 + (35.5x2) + x[ (2x1) + 16]

= 137 + 71 + x[2 + 16]

= 208 + 18x

Mass of BaCl2. xH2O from the question = 1.80g

Number of mole of BaCl2. xH2O =

0.0074 mole

Number of mole = Mass / Molar Mass

0.0074 = 1.80/208 + 18x

Cross multiply to express in linear form as shown below:

0.0074 (208 + 18x) = 1.8

Clear the bracket

1.5392 + 0.1332x = 1.8

Collect like terms

0.1332x = 1.8 - 1.5392

0.1332x = 0.2608

Divide both side by the coefficient of x i. e 0.1332

x = 0.2608/0.1332

x = 2

Therefore the formula for the hydrated salt (BaCl2. xH2O) is BaCl2.2H2O