3. A 3.455-g sample of a mixture was analyzed for barium ion by adding a small excess of sulfuric acid to an aqueous solution of the sample.

Question

3. A 3.455-g sample of a mixture was analyzed for barium ion by adding a small excess of sulfuric acid to an aqueous solution of the sample. The resultant reaction produced a precipitate of barium sulfate, which was collected by filtration, washed, dried, and weighed. If 0.2815 g of barium sulfate was obtained, what was the mass percentage of barium in the sample

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Thu Giang 5 years 2021-07-13T08:38:26+00:00 1 Answers 84 views 0

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  1. bonexptip
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    2021-07-13T08:39:56+00:00
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    Answer:

    Ba\ percentage\ in\ Mass=4.8\%

    Explanation:

    From the question we are told that:

    Mass of mixture m=3.455g

    Mass of Barium m_b=0.2815g

    Equation of Reaction is given as

    Ba2+ + H2SO4 => BaSO4 + 2 H+

    Generally the equation for Moles of Barium  is mathematically given by

    Since

     Moles of Ba^{2+} = Moles of BaSO_4

    Therefore

     Moles of Ba^{2+} = \frac{mass}{molar mass of BaSO4}  

     Moles of Ba^{2+} = \frac{0.2815}{233.39}= 0.0012061 mol

    Generally the equation for Mass of Barium  is mathematically given by

     Mass\ of\ Ba^{2+} = Moles * Molar mass of Ba^{2+}  

     Mass\ of\ Ba^{2+} = 0.0012061 * 137.33 = 0.1656 g

    Therefore

     Ba\ percentage\ in\ Mass = mass of Ba^{2+}/mass of sample * 100%    

     Ba\ percentage\ in\ Mass= \frac{0.1656}{ 3.455 }* 100%

     Ba\ percentage\ in\ Mass=4.8\%

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