How many moles of KF can be produced when 2.67 mol K reacts with 9.01 mol F2?

Initial number of moles of K = 2.67 mol
Initial number of moles of F₂ = 9.01 mol

2K + F₂ → 2KF
Mole ratio K : F₂ = 2 : 1

When K completely reacts,
number of F₂ needed = (2.67 mol) × (1/2) = 1.335 mol < 9.01 mol
Hence, F₂ is in excess, and K completely reacts.

According to the above equation, mole ratio K : KF = 2 : 2 = 1 : 1
Number of moles of K reacted = 2.67 mol
Number of moles of KF produced = 2.67mol

2 K + F2 → 2 KF

2.67 moles of K would react completely with 2.67 x (1/2) = 1.335 moles of F2, but there is more F2 present than that, so F2 is in excess and K is the limiting reactant.

(2.67 mol K) x (2 mol KF / 2 mol K) = 2.67 mol KF

No K is left over because it is the limiting reactant.
(9.01 mol F2 initially) - (1.335 mol F2 reacted) = 7.68 mol F2 left over