Question

A mixture of BaBr2 and inert material is analyzed to determine the Ba content. First the mixture is dissolved in water. Then all of the bromide ion is precipitated as AgBr by the addition of an excess of silver nitrate. BaBr2(aq)——->2AgBr(s)+Ba(No3)2(aq). In one experiment a 0.7207 g sample of the mixture resulted in 0.6226 g of AgBr. Determine the percent (by mass) of Ba in the mixture

Answer 1

We have to calculate the percentage (by mass) of Ba in the mixture of BaBr₂ and inert material.

Given, bromide ion is precipitated as AgBr by the addition of an excess of silver nitrate as per following reaction:

BaBr₂(aq) + 2AgNO₃(aq)→2AgBr(s)+Ba(NO₃)₂(aq). From this balanced chemical reaction, it is clear that one mole BaBr₂ reacts with two moles of AgNO₃ and gives two moles of AgBr. We have to calculate 0.6226 g AgBr contains how many moles of AgBr. Using that information we can get how many moles of BaBr₂ reacted to give 0.6226g AgBr and one mole BaBr₂ contains one mole of Ba. By multiplying number of moles of Ba with atomic mass of Ba we can get amount of Ba present in the mixture.  Accordingly we can calculate mass percentage of Ba in the mixture.  

Atomic mass of Ba= 137.327 g.mol, Molecular mass of BaBr₂=297.1 g/mol and molecular mass of AgBr=187.7 g/mol. Mass of AgBr is 0.6226 g which contains  0.6226/187.7 mole= 3.31X10⁻³ moles of AgBr. So, moles of BaBr₂ reacts= (3.31X10⁻³)/2 moles= 1.65X10⁻³ moles. One mole BaBr₂ contains one mole of Ba. So, 1.65X10⁻³ moles of BaBr₂ contains 1.65X10⁻³ moles of Ba atoms whose mass is=1.65X10⁻³X137.327g=0.2265 g. Out of 0.7207 g sample amount of Ba present is 0.2265 g. So, mass percentage is $\frac{0.2265X100}{0.7207}$=31.42 % .