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
92% (185 ratings)
FREE Expert Solution

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

• gold salt → KAuF4

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
92% (185 ratings)
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.

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. You can view video lessons to learn Limiting Reagent Or if you need more Limiting Reagent practice, you can also practice Limiting Reagent practice problems .

How long does this problem take to solve?

Our expert Chemistry tutor, Kaitlyn took 9 minutes to solve this problem. You can follow their steps in the video explanation above.

What professor is this problem relevant for?

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

What textbook is this problem found in?

Our data indicates that this problem or a close variation was asked in . You can also practice practice problems .