Convert the given masses to moles.
• (55.0 g Na) (1/22.989 mol/g) ≈ 2.39 mol Na
• (67.2 g Cl₂) (1/70.9 mol/g) ≈ 0.948 mol Cl₂
The reaction consumes a ratio of 2 mol Na to 1 mol Cl₂, which means 0.948 mol Cl₂ reacts with 2 × 0.948 ≈ 1.90 mol Na, leaving an excess of about 0.494 mol Na.
The amount of NaCl produced is the same as the amount of Na consumed, so we end up with about 1.90 mol NaCl, which has a mass of about
(1.90 mol NaCl) (58.44 g/mol) ≈ 111 g NaCl
