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?

3 answers

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!