The reaction is,
2 Na + Cl2 → 2 NaCl
( that is, 2 mol of Na react with 1 mol of Cl2 to form 2 mol NaCl.)
— here 1 mol of Na and 1 mol of Cl2 are only there.
.˙. 1 mol of Na react with 0.5 mol of Cl2 to form 1 mol NaCl
Moles of NaCl formed = 1 mol
=> The excess reactant is the Cl2
because only 0.5 mol Cl2 is used.
The other 0.5 mol Cl2 is excess.