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: 3.62b

Answer

Approximately 22.18 grams.

Work Step by Step

1. Convert that number to moles: $mm(O_2) = 16 * 2 = 32$ $n(moles) = \frac{mass(g)}{mm}$ $n(moles) = \frac{7.5}{32}$ $n(moles) \approx 0.234$ 2. Using the proportion given in the problem, find how much moles of $KO_2$ will react with 0.234 moles of $O_2$ $\frac{4}{3} = \frac{x}{0.234}$ $3x = 0.936$ $x = 0.312 moles$ 3. Convert that number to grams: $mm(KO_2) = 39.1*1 + 16*2 = 71.1 g/mol$ $mass(g) = n(moles) * mm$ $mass(g) = 0.312 * 71.1$ $mass(g) \approx 22.18g$
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