Explanation M g ( O H ) 2 : Its reaction with stomach acid is given below. M g ( O H ) 2 ( s ) + 2 H 3 O + ( a q ) → M g 2 + ( a q ) + 4 H 2 O ( l ) One mole of M g ( O H ) 2 is reacting with two moles of stomach acid. The molar mass of M g ( O H ) 2 is 58.32 g / m o l . The amount of M g ( O H ) 2 required for one mole of stomach acid can be calculated as shown below. 58.32 g / m o l 2 = 2 9 . 1 6 g / 1 m o l o f a c i d . C a C O 3 : Its reaction with stomach acid is given below. C a C O 3 ( s ) + 2 H 3 O + ( a q ) → C a 2 + ( a q ) + 3 H 2 O ( l ) + C O 2 ( g ) One mole of C a C O 3 is reacting with two moles of stomach acid. The molar mass of C a C O 3 is 100.10 g / m o l . The amount of C a C O 3 required for one mole of stomach acid can be calculated as shown below. 100.10 g / m o l 2 = 5 0 . 0 5 g / 1 m o l o f a c i d . N a H C O 3 : Its reaction with stomach acid is given below. N a H C O 3 ( s ) + H 3 O + ( a q ) → N a + ( a q ) + 2 H 2 O ( l ) + C O 2 ( g ) One mole of N a H C O 3 is reacting with one moles of stomach acid. The molar mass of N a H C O 3 is 84.00 g / m o l . Thus, the amount of N a H C O 3 required for one mole of stomach acid is 8 4 . 0 0 g / 1 m o l o f a c i d

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ISBN: 9781285199047

Author: John W. Moore, Conrad L. Stanitski

Publisher: Cengage Learning

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Chapter 14.10, Problem 14.13PSP

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The antacid which on a per gram basis, can neutralize the most stomach acid has to be determined.

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Chapter 14.10, Problem 14.28E

Chapter 14.10, Problem 14.29E

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The antacid which on a per gram basis, can neutralize the most stomach acid has to be determined.

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Interpretation: The antacid which on a per gram basis, can neutralize the most stomach acid has to be determined.

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Chapter 14 Solutions

Interpretation:

The antacid which on a per gram basis, can neutralize the most stomach acid has to be determined.