I am completely confused! Help! Redox Reactions!!!?

Question

So my problem says that gold can be oxidized by bromine triflouride. This apparently all happens according to the following redox reaction: 2Au + 2BrF3 + 2KF ---> 2KauF4 + Br2. Say these 3 reactants were all mixed together into equal masses giving it a total mass of 39.2 g. How many grams of KAuF4 were produced from this? Can anyone help me?

Answer 1

Au+BrF3+KF --> KAuF4+1/2Br2
M(Au)=197g/mol; M(BrF3)=136.9g/mol; M(KF)=58.1g/mol
If reactants were all of the same mass and total mass was 39.2g, than it was taken 13.07g of each. In moles it will be
13.07/197=0.66moles of Au
13.07/136.9=0.095 moles of BrF3
13.07/58.1=0.225moles of KF
BrF3 was limiting reagent, since compounds suppose to react in equimolar ratios and the smallest amount of moles taken was of BrF3. According to the main reaction, one mole of KAuF4 is produced per one mole of BrF3 used, so it means that in our reaction we expect to get 0.095 moles of KAuF4, or 29.64g (since M(KAuF4)=312g/mol)

Answer 2

Your statement is a bit vague, I assume you are taking 13.1 grams of each reactant to make a reaction mix. So find the moles of each 13.1 gram component. From the reaction, the ratio of moles of these reactant and the KAuF4 are 1:1. You will find that you can't make more of the product than the lowest moles of any of the reactants. Then you can determine the grams of the product.