Sodium hydrogen carbonate NaHCO3, also known as sodium bicarbonate or "baking soda", can be used to relieve acid indigestion. Acid indigestion is the burning sensation you get in your stomach when it contains too much hydrochloric acid HCl, which the stomach secretes to help digest food. Drinking a glass of water containing dissolved NaHCO3 neutralizes excess HCl through this reaction: HCl(aq)+NaHCO3(aq)→NaCl(aq)+H2O(l)+CO2(g) The CO2 gas produced is what makes you burp after drinking the solution.Suppose the fluid in the stomach of a woman suffering from indigestion can be considered to be 200.mL of a 0.089M HCl solution. What mass of NaHCO3 would she need to ingest to neutralize this much HCl? Be sure your answer has the correct number of significant digits.

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Asked Oct 26, 2019
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Sodium hydrogen carbonate NaHCO3, also known as sodium bicarbonate or "baking soda", can be used to relieve acid indigestion. Acid indigestion is the burning sensation you get in your stomach when it contains too much hydrochloric acid HCl, which the stomach secretes to help digest food. Drinking a glass of water containing dissolved NaHCO3 neutralizes excess HCl through this reaction: HCl(aq)+NaHCO3(aq)→NaCl(aq)+H2O(l)+CO2(g) 

The CO2 gas produced is what makes you burp after drinking the solution.

Suppose the fluid in the stomach of a woman suffering from indigestion can be considered to be 200.mL of a 0.089HCl solution. What mass of NaHCO3 would she need to ingest to neutralize this much HCl? Be sure your answer has the correct number of significant digits.

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Expert Answer

Step 1

Given,

Volume of HCl = 200 mL = 0.200 L    (1 mL = 0.001 L)

Molarity of HCl = 0.089 M = 0.089 mol/L

Moles of HCl can be calculated as:

Moles
Molarity Volume in litres
Moles Molarity x Volume in litres
Moles 0.089 mol/L x 0.200 L
Moles 0.018 mol
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Moles Molarity Volume in litres Moles Molarity x Volume in litres Moles 0.089 mol/L x 0.200 L Moles 0.018 mol

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Step 2

The balanced neutralization reaction of HCl and NaHCO3 is:

HCl (aq) + NaHCO3 (aq) → NaCl (aq) + H2O (l) + CO2 (g)

From the reaction it is evident that:

1 mol HCl requires
1mol NaHCO3
1 mol NaHCO3
:.0.018 mol HCl requires
x 0.018 mol HC
1 mol HCl
= 0.018 mol NaHCO3
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1 mol HCl requires 1mol NaHCO3 1 mol NaHCO3 :.0.018 mol HCl requires x 0.018 mol HC 1 mol HCl = 0.018 mol NaHCO3

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Step 3

Grams of NaHCO3 can be cal...

Molar mass of N2HCO3 23 1 12 (3x16)- 84 g/mol
=
Mass Moles x Molar mass
Mass = 0.018 mol x 84 g/mol
Mass 1.512 g
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Molar mass of N2HCO3 23 1 12 (3x16)- 84 g/mol = Mass Moles x Molar mass Mass = 0.018 mol x 84 g/mol Mass 1.512 g

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