Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 3 - Mass Relationships in Chemical Reactions - Questions & Problems - Page 111: 3.86

Answer

$HCl$ will be used first, producing 23.4 g of $Cl_2$.

Work Step by Step

- Calculate or find the molar mass for $ HCl $: $ HCl $ : ( 35.45 $\times$ 1 )+ ( 1.008 $\times$ 1 )= 36.46 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$ 48.2 \space g \times \frac{1 \space mole}{ 36.46 \space g} = 1.32 \space moles$$ Find the amount of product if each reactant is completely consumed. $$ 0.86 \space mole \space MnO_2 \times \frac{ 1 \space mole \ Cl_2 }{ 1 \space mole \space MnO_2 } = 0.86 \space mole \space Cl_2 $$ $$ 1.32 \space moles \space HCl \times \frac{ 1 \space mole \ Cl_2 }{ 4 \space moles \space HCl } = 0.330 \space mole \space Cl_2 $$ Since the reaction of $ HCl $ produces less $ Cl_2 $ for these quantities, it is the limiting reactant. Therefore, the $HCl$ will be used first, producing 0.330 mole of $Cl_2$. $ Cl_2 $ : ( 35.45 $\times$ 2 )= 70.90 g/mol $$ 0.330 \space mole \times \frac{ 70.90 \space g}{1 \space mole} = 23.4 \space g$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.