   # If 38.55 mL of HCI is required to titrate 2.150 g of Na 2 CO 3 according to the following equation, what is the concentration (mol/L) of the HCl solution? Na 2 CO 3 (aq) + 2HCl(aq) → 2 NaCI(aq) + CO 2 (g) + H 2 O(l) ### 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 71PS
Textbook Problem
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## If 38.55 mL of HCI is required to titrate 2.150 g of Na2CO3 according to the following equation, what is the concentration (mol/L) of the HCl solution?Na2CO3(aq) + 2HCl(aq) → 2 NaCI(aq) + CO2(g) + H2O(l)

Interpretation Introduction

Interpretation:

The concentration (mol/L) of HCl required to titrate 2.150g of Na2CO3 has to be determined.

Concept introduction:

Numberofmole=GivenmassofthesubstanceMolarmass

• 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).
• 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.
• Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.
• Concentrationofsubstance=Amountof substancevolumeofthesubstance

### Explanation of Solution

Balanced chemical equation for the given reaction is,

Na2CO3(aq)+2HCl(aq)H2O(l)+CO2(g)+2NaCl(aq)

The amount (moles) of Na2CO3 available can be calculated as follows,

2.150gNa2CO3×1molNa2CO3106g=0.02028molNa2CO3

The balanced equation for the reaction shows that 1mol of Na2CO3 requires 2mol of HCl.  This is the required stoichiometric factor to obtain the amount of HCl present.

Thus the amount of HCl can be calculated as follows,

0

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