Chapter 4, Problem 70PS

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# What volume of 0.955 M HCl, in milliliters, is required to titrate 2.152 g of Na2CO3 to the equivalence point?Na2CO3(aq) + 2 HCl(aq) → H2O(l) + CO2(g) + NaCl(aq)

Interpretation Introduction

Interpretation:

The volume of 0.955MHCl required to titrate 2.152g of Na2CO3 to the equivalence point has to be calculated in milliliters.

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
• Amountof substance=Concentrationofsubstance×volumeofthesubstance
• Volumeofthesubstance=Amountof substanceConcentrationofsubstance
Explanation

Balanced chemical equation for the given reaction is,

â€‚Â Na2CO3(aq)â€‰+â€‰2HCl(aq)â€‰â†’â€‰H2O(l)â€‰+â€‰CO2(g)+â€‰2â€‰NaCl(aq)

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

â€‚Â 2.152â€‰gâ€‰Na2CO3â€‰Ã—1â€‰molâ€‰Na2CO3106â€‰gâ€‰â€‰=â€‰0.02030â€‰molâ€‰Na2CO3

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

Thus the amount of HCl can be calculated as follows,

â€‚Â 0

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