Problem: Sodium and chlorine react to form sodium chloride:2 Na(s) + Cl2(g) → 2NaCl(s)What is the theoretical yield of sodium chloride for the reaction of 55.0 g Na with 67.2 g Cl2?

2 Na(s) + Cl₂(g) → 2 NaCl(s)

Calculate the amount of NaCl produced from each reactant and determine the limiting reactant:

mass Na (molar mass) → moles Na (mole-to-mole comparison) → moles NaCl (molar mass) → mass NaCl

molar mass Na = 23.0 g/mol
molar mass NaCl = 58.45 g/mol

$\mathbf{mass}\mathbf{ }\mathbf{NaCl}\mathbf{=}\mathbf{55}\mathbf{.}\mathbf{0}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }\mathbf{Na}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }\mathbf{Na}}}{\mathbf{23}\mathbf{.}\mathbf{0}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }\mathbf{Na}}}\mathbf{\times }\frac{\mathbf{2}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }\mathbf{NaCl}}}{\mathbf{2}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }\mathbf{Na}}}\mathbf{\times }\frac{\mathbf{58}\mathbf{.}\mathbf{45}\mathbf{ }\mathbf{g}\mathbf{ }\mathbf{NaCl}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }\mathbf{NaCl}}}$

mass NaCl = 139.77 g NaCl

mass Cl₂ (molar mass) → moles Cl₂ (mole-to-mole comparison) → moles NaCl (molar mass) → mass NaCl