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 73.0 -g mixture of equal masses of all three reactants is prepared.

this was a little helpful but no exactally what i was looking for

Coool. Beans

Well, well, well, it seems like we have some chemistry enthusiasts here! Alright, let's solve this golden question together.

First, we need to find the molar mass of each reactant:
BrF3 (Bromine Trifluoride) = 127.9 g/mol
KF (Potassium Fluoride) = 58.1 g/mol
KAuF4 (Potassium Gold Tetrafluoride) = 285.0 g/mol

Since the 73.0 g mixture consists of equal masses of all three reactants, each reactant will have a mass of 73.0 g / 3 = 24.3 g.

Now, let's find how many moles of each reactant we have:
moles of BrF3 = mass / molar mass = 24.3 g / 127.9 g/mol = 0.1902 mol
moles of KF = mass / molar mass = 24.3 g / 58.1 g/mol = 0.4187 mol

The reaction between BrF3 and gold metal tells us that 2 moles of BrF3 react with 1 mole of gold to produce 1 mole of KAuF4. Therefore, the limiting reactant is BrF3 since it has fewer moles than KF.

Now we need to find the number of moles of Br2 that can form:
moles of Br2 = moles of BrF3 x (1 mole of Br2 / 2 moles of BrF3) = 0.1902 mol x (1 mol / 2 mol) = 0.0951 mol

Finally, we can find the mass of the gold salt (KAuF4) that forms using its molar mass:
mass of KAuF4 = moles of KAuF4 x molar mass of KAuF4
Let's plug in the values:
mass of KAuF4 = 0.0951 mol x 285.0 g/mol = 27.1395 g

So, the mass of the gold salt that forms when a 73.0 g mixture of equal masses of all the reactants is prepared is approximately 27.1 g.

Remember, chemistry jokes might not get a reaction, but careful calculations definitely will!

To find the mass of the gold salt (KAuF4) that forms when a 73.0 g mixture of equal masses of all three reactants is prepared, we need to use stoichiometry and the balanced chemical equation for the reaction.

The balanced chemical equation for the reaction is:
2 Au + 5 BrF3 + 10 KF → 2 KAuF4 + 5 Br2

From the balanced equation, we can see that 2 moles of gold (Au) react with 5 moles of BrF3 and 10 moles of KF to produce 2 moles of KAuF4 and 5 moles of Br2.

Step 1: Calculate the molar masses of the reactants and the product:
- Au: 196.97 g/mol
- BrF3: 136.90 g/mol
- KF: 58.10 g/mol
- KAuF4: The molar mass of KAuF4 can be calculated by adding the molar masses of potassium (K), gold (Au), and fluorine (F). The molar mass of KAuF4 is 272.0 g/mol.

Step 2: Convert the mass of the mixture to moles:
molar mass of the mixture = (molar mass of Au + molar mass of BrF3 + molar mass of KF) / 3
= (196.97 g/mol + 136.90 g/mol + 58.10 g/mol) / 3
= 130.99 g/mol

moles of the mixture = mass of the mixture / molar mass of the mixture
= 73.0 g / 130.99 g/mol
≈ 0.557 mol

Step 3: Use stoichiometry to determine the moles of KAuF4 produced:
From the balanced equation, we can see that the moles of KAuF4 produced are the same as the moles of Au used.
moles of KAuF4 = 2 * moles of Au
= 2 * 0.557 mol
= 1.114 mol

Step 4: Convert moles of KAuF4 to mass:
mass of KAuF4 = moles of KAuF4 * molar mass of KAuF4
= 1.114 mol * 272.0 g/mol
≈ 302.61 g

Therefore, the mass of the gold salt (KAuF4) that forms when a 73.0 g mixture of equal masses of all three reactants is prepared is approximately 302.61 g.

equal masses means 73.0/3 = 24.33 g each.

Calculate moles each. moles = grams/molar mass.
Using the coefficients in th balanced equation, convert each to moles of the salt. The smallest value will be the amount of salt that forms and that reagent will be the limiting reagent. Use that value to convert to grams. g = moles x molar mass.