2 Na(s) + Cl_{2}(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

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

**mass NaCl = 139.77 g NaCl**

mass Cl_{2} (molar mass) → moles Cl_{2} (mole-to-mole comparison) → moles NaCl (molar mass) → mass NaCl

Sodium and chlorine react to form sodium chloride:

2 Na(s) + Cl_{2}(g) → 2NaCl(s)

What is the theoretical yield of sodium chloride for the reaction of 55.0 g Na with 67.2 g Cl_{2}?

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