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.

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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
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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.

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