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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okpalawalter8 okpalawalter8 · Answer 1 · 2021-07-01T21:40:46+02:00

Answer:

[tex]Ba\ percentage\ in\ Mass=4.8\%[/tex]

Explanation:

From the question we are told that:

Mass of mixture [tex]m=3.455g[/tex]

Mass of Barium [tex]m_b=0.2815g[/tex]

Equation of Reaction is given as

[tex]Ba2+ + H2SO4 => BaSO4 + 2 H+[/tex]

Generally the equation for Moles of Barium is mathematically given by

Since

[tex]Moles of Ba^{2+} = Moles of BaSO_4[/tex]

Therefore

[tex]Moles of Ba^{2+} = \frac{mass}{molar mass of BaSO4}[/tex]

[tex]Moles of Ba^{2+} = \frac{0.2815}{233.39}= 0.0012061 mol[/tex]

Generally the equation for Mass of Barium is mathematically given by

[tex]Mass\ of\ Ba^{2+} = Moles * Molar mass of Ba^{2+}[/tex]

[tex]Mass\ of\ Ba^{2+} = 0.0012061 * 137.33 = 0.1656 g[/tex]

Therefore

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

[tex]Ba\ percentage\ in\ Mass= \frac{0.1656}{ 3.455 }* 100%[/tex]

[tex]Ba\ percentage\ in\ Mass=4.8\%[/tex]