If 64.9 g of sodium and 62.3 g of chlorine are used, how many grams of sodium chloride are produced?

Molar mass of Na = 22.99 g/mol
Molar mass of Cl₂ = 35.45 × 2 g/mol = 70.90 g/mol
Molar mass of NaCl = (22.99 + 35.45) g/mol = 58.44 g/mol

2Na + Cl₂ → 2NaCl
OR: Mole ratio Na : Cl₂ : NaCl = 2 : 1 : 2

If 64.9 g of sodium reacts completely :
No. of moles of Na reacted = (64.9 g) /(22.99 g/mol) = 64.9/22.99 mol
Mass of NaCl produced = (64.9/22.99 mol) × (2/2) × (58.44 g/mol) = 165 g

If 62.3 g of chlorine reacts completely :
No. of moles of Cl₂ reacted = (62.3 g) / (70.90 g/mol) = 62.3/70.90 mol
Mass of NaCl produced = (62.3/70.90 mol) × 2 × (58.44 g/mol) = 103 g

The limiting reactant would give a smaller amount of product.
Hence, chlorine is the limiting reactant.
Mass of NaCl produced = 103 g

64.9*(35.5/23) ~ 100.2g with a total amount of leftover chlorine of about 27g.