Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 4 - Sections 4.1-4.9 - Exercises - Problems by Topic: 44


$0.206$ $mol$ $HCl$

Work Step by Step

Determine the limiting reactant. Choose one of the reactants and determine how much we need compared to how much is available. We choose to compare HCl. $0.652 mols$ $HCl (avaliable)$ $0.223 mols$ $FeS$$\times\frac{2 mol HCl}{1 mol FeS}$$=0.446$ $mols$ $HCl (needed)$ The moles of excess reactant is found by subtracting the needed amount from the available amount. Excess amount of $HCl$ = $avaliable - needed$ =$0.652$ $mols$ - $0.446$ $mols$ $HCl$ =$0.206$ $mols$ $HCl$
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