Find the mass of the gold salt?

Question

Hello, I'm stuck on this question, can someone help me out... is there some magical formula to calculate this:

Au is treated with Br3 and KF and produces KAuF4 gold salt plus Br2.

Calculate the mass of gold salt - forms when 76.0g mixture of equal masses of all three reactants is prepared.

Thank you very much!

Answer 1

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.