Answer
$66.5\ kg$
Work Step by Step
Number of moles of $CaCO_3$:
$125.\ kg\times 95.0/100\div 100.09\ kg/kmol= 1.186\ kmol$
Mass of lime:
$1.186\ kmol\times 56.08\ kg/kmol=66.5\ kg$
Chemistry and Chemical Reactivity (9th Edition)
by Kotz, John C.; Treichel, Paul M.; Townsend, John R.; Treichel, David A.
Published by
Cengage Learning
ISBN 10:
1133949649
ISBN 13:
978-1-13394-964-0
Chapter 4 Stoichiometry: Quantitative Information about Chemical Reactions - Study Questions - Page 179h: 99
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