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

Answer 1

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

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

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

 $Ba\ percentage\ in\ Mass=4.8\%$