Answer
$n_{CaCO_3}=0.1854\ mol$
Work Step by Step
$K_p=p(CO_2)$
$p(CO_2)=3.87\ atm$
$n_{CO_2}=n_{CaCO_3}=p(CO_2)V/RT$
$n_{CaCO_3}=3.87\ atm\times 5.00\ L/(0.082\ L.atm/mol.K\times1273\ K)$
$n_{CaCO_3}=0.1854\ mol$
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 15 Principles of Chemical Reactivity: Equilibria - Study Questions - Page 583f: 58
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