Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 3 - Chemical Reactions and Reaction Stoichiometry - Exercises - Page 116: 3.63b


0.701 g de HCl

Work Step by Step

First, we calculate the molecular mass of $Al(OH)_3$. Molecular mass of $Al(OH)_3$= (1x27)+(3x16)+(3x1) = 78 g/mol. How many moles of $Al(OH)_3$ have in 0.500 g? $\frac{0,500 g}{78g/mol}= 6,41\times10^{-6}$ mols de $Al(OH)_3$ This will react with $6,41\times10^{-6}\times3$ HCl = 0,0192 mol HCl. Now, we will find the molar mass of HCl: MM(HCl) = (1x1,008)+(1x35,45) = 36,5 g/mol Finally, the mass of 0.0192 mols = 0,0192 mols x 36,5 g/mol = 0,701 g HCl.
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