Treatment of gold metal with BrF3 and KF produces Br2 and KAuF4, a salt of gold.

Question

Treatment of gold metal with BrF3 and KF produces Br2 and KAuF4, a salt of gold. Write a balance chemical equation and identify the oxidizing agent and reducing agent in the reaction. Find the mass of gold salt that forms when a 73.5g mixture of equal masses of all three reactants is prepared.

Explanation / Answer

There is no compound Br3. Br3- is called tribromide and is made from mixing Br2 with aqueous Bromide solutions. Are you sure that the Gold salt was not made with BrF3 (Bromine trifluoride) and KF (Potassium fluoride)? 2 Au + 2 BrF3 + 2 KF ---> 2 KAuF4 + Br2 One mole of each reactant is need to form one mole of KAuF4 You need to calculate how many moles of each reactant are in 25.33 grams (76.0 g / 3). Au 25.33 g / 196.97 g/mol = 0.1285 moles BrF3 25.33 g / 136.90 g/mol = 0.1850 moles KF 25.33 g / 58.10 g/mol = 0.4360 moles The Gold is the limiting chemical. To get your answer, multiply 0.1285 moles by the Molecular mass of the KAuF4.

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