Reactions involving gold have been of particular interest to alchemists. Consider the following reactions,
$Au (OH)_{3} + 4 HCl ⟶ {HAuCl}_{4} + 3 H_{2} O$ , $Δ H = - 28 kcal$
$Au (OH)_{3} + 4 HBr ⟶ {HAuBr}_{4} + 3 H_{2} O$ , $Δ H = - 36.8 kcal$
In an experiment there was an absorption of $0.44 kcal$ when one mole of ${HAuBr}_{4}$ was mixed with 4 moles of $HCl$ . Then the fraction ${HAuBr}_{4}$ converted into ${HAuCl}_{4}$ : (percentage conversion)

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Detailed Solution

The general formula is

${HAuBr}_{4} + 4 HCl ⟶ {HAuCl}_{4} + 4 HBr$

Then

$Δ H = - 28 + 36.8 = 8.8$

$\frac{0.44}{8.8} \times 100 = 5 %$

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