Treatment of gold metal with BrF₃ and KF produces Br₂ and KAuF₄, a salt of gold. Find the mass of the gold salt that forms when a 73.5-g mixture of equal masses of all three reactants is prepared.

Video transcript

Hi everyone today. We have a question telling us when golden idol is treated with roaming. Try fluoride and potassium fluoride bro. Mean and potassium gold florid are produced 55.2 g mixture of equal masses of all three reactant is prepared. What is the mass of the browning that is produced from the mixture? So first we need a balanced reaction. So we have gold plus bro. Ming trifle ride plus potassium fluoride forms bro, ming and potassium gold fluoride. So we're going to start here by balancing out our bro. Mean by putting a two here and then we'll balance out our flooring by putting a two here and then we'll balance our gold by putting a two here. And lastly we'll balance our potassium by putting a two here. So now that we've got our balanced chemical equation, let's look at our molar masses. So our molar masses. We have gold Which is 196.97 g per mole. We have bro ming try fluoride, Which is 136.90 g per mole. We have potassium fluoride Which is 58 .10 g Permal. Now we have the same amount of all three of these. We have 55.2 g Divided by three, which gives us 18.4 g. So the elements with the highest number of Mueller mass is going to give us the least number of moles, which in this case is gold. So that is going to be our limiting a re agent are limiting reactant. So now we can solve for the massive bruning. We have 18.4 grams of gold times one mole of gold Over 146, .97 g of gold times one mole of roaming over two moles of gold Times 159.8 grams of roaming over one mole of browning. And our grams of gold are going to cancel out. Our moles of gold are going to cancel out and our moles of roaming are going to cancel out. And that is going to equal 7.46 g of roaming. And that is our final answer. Thank you for watching. Bye.