111 grams
First, look up the atomic weights of sodium and chlorine.
Atomic weight sodium = 22.989769
Atomic weight chlorine = 35.453
Molar mass Cl2 = 2 * 35.453 = 70.906 g/mol
Molar mass NaCl = 22.989769 + 35.453 = 58.442769 g/mol
Now determine how many moles of each reactant you have.
Moles Na = 55.0 g / 22.989769 g/mol = 2.392368536 mol
Moles Cl2 = 67.2 g / 70.906 g/mol = 0.947733619 mol
Looking at the balanced equation, it takes 2 moles of Na per mole of Cl2. Given the number of respective moles, it's obvious that Cl2 is the limiting reactant in this problem. And looking at the balanced equation, for every mole of Cl2 consumed, 2 moles of NaCl is produced. So do the multiplication.
2 * 0.947733619 mol = 1.895467238 mol
And multiply by the molar mass, so
1.895467238 * 58.442769 = 110.776354 g
Round to 3 significant figures, giving 111 g
