Problem: Treatment of gold metal with BrF3 and KF produces Br2 and KAuF4, a salt of gold. If a 77.5 g mixture of equal masses is used, find the actual mass of the gold salt that forms.

FREE Expert Solution

Balanced equation:  2Au + 2BrF+ 2KF → Br2 + 2KAuF4

  • gold salt → KAuF4


mass of each reactant = mass of mixture3=77.5 g3=25.83 g


Determine the limiting reactant → reactant that forms the less amount of 

product


Comparing molar mass: 

  • Molar Mass Au = 196.97 g/mol
  • Molar Mass BrF= 136.90 g/mol
  • Molar Mass KF = 58.10 g/mol
    • since equal masses:
      • Au will have least # of moles → forms least amt. of product  limiting reactant  
        • 2 moles of each reactant produces 2 moles of gold salt 


The mass of gold salt formed by 25.83 g Au is:

  • Molar Mass KAuF= 312.06 g/mol
View Complete Written Solution
Problem Details

Treatment of gold metal with BrF3 and KF produces Br2 and KAuF4, a salt of gold. If a 77.5 g mixture of equal masses is used, find the actual mass of the gold salt that forms.

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Limiting Reagent concept. If you need more Limiting Reagent practice, you can also practice Limiting Reagent practice problems.

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Iob's class at OKSTATE.