Chemistry and Chemical Reactivity (9th Edition)

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 179i: 113


$1.73\ g$

Work Step by Step

Number of moles of $CaCO_3$: $2.56\ g\div 100.09\ g/mol=0.0256\ mol$ Number of moles of $HCl$: $0.125\ M\times 0.250\ L=0.0313\ mol$ Ratio: $1.22$, HCl is the limiting reactant, so calcium carbonate remains after the reaction is complete. Mass of calcium chloride: $0.0313\ mol \times 1/2\times 110.98\ g/mol=1.73\ g$
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