General Chemistry 10th Edition

Published by Cengage Learning
ISBN 10: 1-28505-137-8
ISBN 13: 978-1-28505-137-6

Chapter 3 - Calculations with Chemical Formulas and Equations - Exercises - Page 112: 3.17

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.
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