Answer
HCl gives the smallest amount of $AlCl_{3}$, so this substance is the limiting reactant. The molesof $AlCl_{3}$ that can be obtained are 0.12 mol.
Work Step by Step
Strategy: Find how many moles AlCl3 yield the given moles of each reactant.
The one reactant that yields the smaller amount of product is the limiting reactant.
2 Al (s) + 6 HCl (g) ${\longrightarrow}$ 2Al$Cl_{3}$ (s) + 3$H_{2}$ (g)
Step 1:
0.15 mol Al x $\frac{2 mol(AlCl_{3})}{2 mol(Al)}$= 0.15 mol Al$Cl_{3}$
0.35 mol HCl x $\frac{2 mol(AlCl_{3})}{6 mol(HCl)}$ = 0.12 mol $AlCl_{3}$
Step 2: 0.35 mol HCl gives the smallest amount of $AlCl_{3}$, so this substance is the limiting reactant. The molesof $AlCl_{3}$ that can be obtained are 0.12 mol.