Question

A mixture consisting of only iron(III) bromide (FeBr3) and aluminum bromide (AlBr3) weighs 0.9553 g. When the mixture is dissolved in water and an excess of silver nitrate is added, all the bromide ions associated with the original mixture are precipitated as insoluble silver bromide (AgBr). The mass of the silver bromide is found to be 1.9182 g. Calculate the mass percentages of iron(III) bromide and aluminum bromide in the original mixture.

Answer 1

In this answer, use RAM (Fe=55.9, Ag=107.9, Al=27.0, N=14, )=16, Br=79.9)

Step I: Balanced equation for the reaction

AlBr_3(aq) + FeBr_3(aq) + 6AgNO_3(aq) $\to$ 6AgBr_(s) + Al(NO₃)_3(aq) + Fe(NO₃)_3(aq)

Step II: Moles of the product and the reactants

RFM of AgBr = 107.9 + 79.9 = 187.8g/mol

That means 1 mole of AgBr = 187.8g

? moles = 1.9182g

$= \dfrac{1molx1.9182g}{187.8g} = 0.0102moles$

Mole ratio of AgBr:AlBr₃ = 6:1

Moles of AlBr₃ $=$ $\dfrac{1}{6}x0.0102 = 0.0017moles$

Likewise, moles of FeBr₃$=\dfrac{1}{6}x0.0102=0.0017moles$

Step III: Mass of the reactants

RFM of AlBr₃ = 27 + (3x79.9) = 266.7g/mol

That means 1mol of AlB₃ =266.7g

0.0017moles = ?

$=\dfrac{0.0017mol x 266.7g}{1mol}= 0.45339g$

% by mass of AlBr₃ in the original mixture$= \dfrac{0.4533g x100}{0.9553g} = 47.46$ %

RFM of FeBr₃ = 55.9 + (3 x 79.9) = 295.6g/mol

That means 1 mole of FeBr3=295.6g

0.0017moles = ?

$=\dfrac{0.0017molx 295.6g}{1mol}=0.50252g$

% by mass of FeBr₃ in the original sample =$\dfrac{0.50252gx100}{0.9553g}=52.6$ %