How many grams of the excess reagent are left over if 37.8g of cl2 react with 39.4 g of naf?

The reaction is as follows:

2 NaF + Cl₂ → 2 NaCl + F₂

Moles NaF = 39.4 g/ 42 g/mol = 0.938 mol
Moles Cl₂ = 37.8 g/ 70.9 g/mol = 0.533 mol

Find the theoretical amount of Cl₂ needed for the given NaF:
0.938 mol NaF (1 mol Cl₂/2 mol NaF) = 0.469 mol Cl₂

Thus, the excess reactant is Cl₂. When all of the limiting reactant, NaF, is consumed, the amount of excess reactant left would be:
Excess Reagent = (0.533 mol - 0.469 mol) * 70.9 g/mol = 4.54 g Cl₂

Ci think not