   # Sodium bicarbonate and acetic acid react according to the equation NaHCO 3 (aq) + CH 3 CO 2 H(aq) → NaCH 3 CO 2 (aq) + CO 2 (g) + H 2 O(l) What mass of sodium acetate can be obtained from mixing 15.0 g of NaHCO 3 with 125 mL of 0.15 M acetic acid? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 4, Problem 105GQ
Textbook Problem
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## Sodium bicarbonate and acetic acid react according to the equationNaHCO3(aq) + CH3CO2H(aq) → NaCH3CO2(aq) + CO2(g) + H2O(l)What mass of sodium acetate can be obtained from mixing 15.0 g of NaHCO3 with 125 mL of 0.15 M acetic acid?

Interpretation Introduction

Interpretation:

The mass of sodium acetate obtained in the given reaction has to be determined.

Concept introduction:

• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.
• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.

Concentrationofsubstance=Amountof substancevolumeofthesubstance

• Amountof substance=Concentrationofsubstance×volumeofthesubstance
• Limiting reagent: Limiting reagent is a reactant, which consumes completely in the chemical reaction.  The quantity of the product depends on this limiting reagent, it can be determined with the help of balanced chemical equation for the reaction

### Explanation of Solution

Balanced chemical equation for the given reaction is,

NaHCO3(aq)+CH3CO2H(aq)NaCH3CO2(aq)+CO2(g)+H2O(l)

From the balanced chemical equation it is clear that reactants and products react in a 1:1 ratio.

The amount (moles) of CH3CO2H used in the reaction can be calculated from its volume and concentration as follows,

AmountofCH3CO2H =CCH3CO2H×VCH3CO2H =0.15molL×0.125L =0.01875molCH3CO2H

Using the stoichiometric factor the amount of NaCH3CO2 can be calculated as follows,

0

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