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

`Ba\ percentage\ in\ Mass=4.8\%`

Step-by-step 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+) = (mass)/(molar mass of BaSO4)`

`Moles of Ba^(2+) = (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= (0.1656)/( 3.455 )* 100%`

`Ba\ percentage\ in\ Mass=4.8\%`