What mass of potassium chloride forms when 5.11 L of chlorine gas at 0.943 atm and 286 K reacts with 29.0 g potassium?

Consider the Cl₂ gas.
Gas law: PV = nRT
Then, V = PV/(RT)
Initial number of moles of Cl₂, n = 0.943 × 5.11 / (0.08206 × 286) mol = 0.2053 mol

Molar mass of K = 39.0 g/mol
Initial number of moles of K = (29.0 g) / (39.0 g/mol) = 0.7436 mol

Balanced equation for the reaction :
2K + Cl₂ → 2KCl
Mole ratio K : Cl₂ = 2 : 1

When 0.2053 mol of Cl₂ completely reacts, K needed = (0.2053 mol) × 2 = 0.4106 mol < 0.7436 mol
Hence, K is in excess, while Cl₂ completely reacts.

According to the above equation, mole ratio Cl₂ : KCl = 1 : 2
Number of moles of Cl₂ (completely) reacted = 0.2053 mol
Number of moles of KCl formed = (0.2053 mol) × 2 = 0.4106 mol

Molar mass of KCl = (39.0 + 35.5) g/mol = 74.5 g/mol
Mass of KCl formed = (0.4106 mol) × (74.5 g/mol) = 30.6 g