CHM 111/112 LAB MANUAL >C<
CHM 111/112 LAB MANUAL >C<
11th Edition
ISBN: 9781337310956
Author: SLOWINSKI
Publisher: CENGAGE LEARNING (CUSTOM)
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Chapter 28, Problem 1ASA

(a)

Interpretation Introduction

Interpretation : The moles of CaCO3 used during the reaction of 0.3946 g of sample of CaCO3 with 12M HCl to form the final solution as 250 mL should be determined.

Concept Introduction : The mole concept is mainly used to determine the stoichiometry of any chemical reaction. It is used to calculate the mass of the reactant used or the amount of product formed during the chemical reaction. The relation between mass and moles of any chemical substance can be written as:

  Moles=massgmolar massg/mol

(a)

Expert Solution
Check Mark

Answer to Problem 1ASA

In the given reaction, 0.0039 moles of CaCO3 are used.

Explanation of Solution

Given:

Mass of CaCO3 = 0.3946 g

Formula mass of CaCO3 = 100.1 u

Concentration of HCl = 12 M

Final volume of solution = 250 mL

Substitute the values in the given formula to calculate no. of moles of CaCO3 as shown:

  Moles=massgmolar massg/mol

  No. of moles of CaCO3=0.3946 g100.1 g/mol=0.0039 moles

(b)

Interpretation Introduction

Interpretation : The molarity of Ca2+ in 250 mL should be determined during the reaction of 0.3946 g of sample of CaCO3 with 12 M HCl .

Concept Introduction : The mole concept is mainly used to determine the stoichiometry of any chemical reaction. It is used to calculate the mass of the reactant used or the amount of product formed during the chemical reaction. The relation between mass and moles of any chemical substance can be written as:

  Moles=massgmolar massg/mol

(b)

Expert Solution
Check Mark

Answer to Problem 1ASA

In the given reaction the molarity of Ca2+ ions must be 0.0156 M

Explanation of Solution

Given:

Mass of CaCO3 = 0.3946 g

Formula mass of CaCO3 = 100.1 u

Concentration of  HCl = 12 M

Final volume of solution = 250 mL

Calculated the moles of CaCO3 = 0.0039 moles

Expression for molarity is as follows:

  Molarity=No. of molesVolume in L

Hence molarity of Ca2+ ions must be = 0.0039 moles250 mL×1000 mL1 L=0.0156 M

(c)

Interpretation Introduction

Interpretation : The mole of Ca2+ in 25.00 mL should be determined during the reaction of 0.3946 g of sample of CaCO3 with 12 M HCl .

Concept Introduction : The mole concept is mainly used to determine the stoichiometry of any chemical reaction. It is used to calculate the mass of the reactant used or the amount of product formed during the chemical reaction. The relation between mass and moles of any chemical substance can be written as:

  Moles=massgmolar massg/mol

(c)

Expert Solution
Check Mark

Answer to Problem 1ASA

In the given reaction mole of Ca2+ in 25.00 mL should be 0.00039 moles .

Explanation of Solution

Given:

Molarity of Ca2+ ions = 0.0156 M

Volume = 25.00 mL

Relation between molarity and volume is as follows:

  Moles=molarity×volume

Substitute values in the above expression to calculate no. of moles as follows:

  Moles=molarity×volume=0.0156 molL×0.025 L=0.00039 moles

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