BuyFind

Chemistry for Today: General, Orga...

9th Edition
Spencer L. Seager + 2 others
Publisher: Cengage Learning
ISBN: 9781305960060
BuyFind

Chemistry for Today: General, Orga...

9th Edition
Spencer L. Seager + 2 others
Publisher: Cengage Learning
ISBN: 9781305960060

Solutions

Chapter 7, Problem 7.60E
Textbook Problem

Stomach acid is essentially 0.10 M HCl . An active ingredient found in a number of popular antacids is calcium carbonate, CaCO 3 . Calculate the number of grams of CaCO 3 needed to exactly react with 250. mL of stomach acid.

CaCO 3 ( s ) + 2 HCl ( aq ) CO 2 ( g ) + CaCl 2 ( aq ) + H 2 O ( l )

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Chapter 7 Solutions

Chemistry for Today: General, Organic, and Biochemistry
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