Answer
NaOH gives the smallest amount of NaClO, so this substance is the limiting reactant. The molesof NaClO that can be obtained is 0.72 mol.
Work Step by Step
Strategy: Find how many moles NaClO yield the given moles of each reactant.
The one reactant that yields the smaller amount of product is the limiting reactant.
2 NaOH (aq) + $Cl_{2}$ ${\longrightarrow}$ NaCl (aq) + NaClO (aq) + $H_{2}O$ (l)
Step 1:
1.44 mol NaOH x $\frac{1 mol(NaClO)}{2 mol(NaOH)}$= 0.72 mol NaClO
1.47 mol $Cl_{2}$ x $\frac{1 mol(NaClO)}{1 mol(Cl_{2})}$ = 1.47 mol NaClO
1.44 mol NaOH gives the smallest amount of NaClO, so this substance is the limiting reactant. The molesof NaClO that can be obtained is 0.72 mol.
