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28 July, 03:45

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.

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  1. 28 July, 03:52
    0
    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
  2. 28 July, 03:53
    0
    X = 1553 benacne vamsy ue etanbe
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