PF3 reacts with XeF4 to give PF5

2PF3(g) + XeF4(s) --> 2PF5(g) + Xe(g)

How many moles of PF5 can be produced from 100.0 grams of PF3 and 50.0 grams of XeF4?

100.0gPF3 x (1molePF3 / 87.96896gPF3) x (2molesPF5 / 2molesPF3) = 1.14molesPF5

50.0gXeF4 x (1moleXeF4 / 207.2836gPF3) x (2molePF5 / 1moleXeF4) = .482moles PF5

I know the answer is .482molesPF5. Is that because XeF4 is the limiting reactant?

I think you need some help on limiting reagents and finding mols and stoichiometry.

Here is step by step but no numbers.

1. You have the balanced equation.
2a. Convert grams PF3 to mols. mols = grams/molar mass = ?
2b. Do the same to convert mols XeF4 to mols PF5.

3a. Using the coefficients in the balanced equation, convert mols PF3 to mols PF5.
3b. Do the same and convert mols XeF4 to mols PF5.
3c. It is likely the two values for mols PF5 will not agree which means one of them is not right. The correct value in limiting reagent problems is ALWAYS the smaller number.

4. The problem doesn't ask for it but if you want to convert mols PF5 to grams, that is grams PF5 = mols x molar mass = ?

Thanks!

1g

Yes, the reason why the answer is 0.482 moles of PF5 is because XeF4 is the limiting reactant in this reaction.

To determine the limiting reactant, you compare the number of moles of each reactant with their stoichiometric coefficients. In this case, you have 100.0 grams of PF3 and 50.0 grams of XeF4.

First, calculate the number of moles of each reactant:
- Moles of PF3 = 100.0g / molar mass of PF3 = 100.0g / 87.96896g/mol = 1.14 moles
- Moles of XeF4 = 50.0g / molar mass of XeF4 = 50.0g / 207.2836g/mol = 0.241 moles

Next, use the stoichiometry of the balanced equation to determine the moles of PF5 that can be produced from each reactant:
- From 1 mole of PF3, you can produce 2 moles of PF5.
- From 1 mole of XeF4, you can also produce 2 moles of PF5.

Calculate the moles of PF5 that can be produced from each reactant:
- Moles of PF5 from PF3 = 1.14 moles PF3 * (2 moles PF5 / 2 moles PF3) = 1.14 moles PF5
- Moles of PF5 from XeF4 = 0.241 moles XeF4 * (2 moles PF5 / 1 mole XeF4) = 0.482 moles PF5

Since you have fewer moles of PF5 from XeF4, it is the limiting reactant in this reaction. Therefore, you can only produce 0.482 moles of PF5 from the given amounts of reactants.