# 5.6 mols of sodium reacts with 3.4 mols of chlorine. How many mols of sodium chloride are formed? What is the limiting reactant?

Mar 14, 2016

$N a \left(s\right) + \frac{1}{2} C {l}_{2} \left(g\right) \rightarrow N a C l \left(s\right)$
We assume standard conditions. Thus, there are 5.6 mol sodium metal and 3.4 mols $C {l}_{2}$ gas, that is 6.8 mol chlorine atoms . Thus the metal is the reagent in deficiency, and we can form 5.6 mol of salt.