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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]