Chapter 28, Problem 1ASA

(a)

Interpretation Introduction

Interpretation : The moles of ${\text{CaCO}}_{\text{3}}$ used during the reaction of $0.3946\text{ g}$ of sample of ${\text{CaCO}}_{\text{3}}$ with $12M\text{ HCl}$ to form the final solution as $250\text{ mL}$ 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:

  $\text{Moles}=\frac{\text{mass}\left(\text{g}\right)}{\text{molar mass}\left(\text{g}/\text{mol}\right)}$

Answer to Problem 1ASA

In the given reaction, $0.0039\text{ moles}$ of ${\text{CaCO}}_{\text{3}}$ are used.

Explanation of Solution

Given:

Mass of ${\text{CaCO}}_{\text{3}}$ = $0.3946\text{ g}$

Formula mass of ${\text{CaCO}}_{\text{3}}$ = $100.1\text{ u}$

Concentration of $\text{HCl}$ = $12\text{ M}$

Final volume of solution = $250\text{ mL}$

Substitute the values in the given formula to calculate no. of moles of ${\text{CaCO}}_{\text{3}}$ as shown:

  $\text{Moles}=\frac{\text{mass}\left(\text{g}\right)}{\text{molar mass}\left(\text{g}/\text{mol}\right)}$

  $\begin{array}{c}\text{No}{\text{. of moles of CaCO}}_{\text{3}}=\frac{0.3946\text{ g}}{100.1\text{g}/\text{mol}}\\ =0.0039\text{ moles}\end{array}$

(b)

Interpretation Introduction

Interpretation : The molarity of ${\text{Ca}}^{2+}$ in $250\text{ mL}$ should be determined during the reaction of $0.3946\text{ g}$ of sample of ${\text{CaCO}}_{\text{3}}$ with $12\text{ M HCl}$ .

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:

  $\text{Moles}=\frac{\text{mass}\left(\text{g}\right)}{\text{molar mass}\left(\text{g}/\text{mol}\right)}$

Answer to Problem 1ASA

In the given reaction the molarity of ${\text{Ca}}^{2+}$ ions must be $0.0156\text{ M}$

Explanation of Solution

Given:

Mass of ${\text{CaCO}}_{\text{3}}$ = $0.3946\text{ g}$

Formula mass of ${\text{CaCO}}_{\text{3}}$ = $100.1\text{ u}$

Concentration of $\text{ HCl}$ = $12\text{ M}$

Final volume of solution = $250\text{ mL}$

Calculated the moles of ${\text{CaCO}}_{\text{3}}$ = $0.0039\text{ moles}$

Expression for molarity is as follows:

  $\text{Molarity}=\frac{\text{No}\text{. of moles}}{\text{Volume in L}}$

Hence molarity of ${\text{Ca}}^{2+}$ ions must be = $\frac{0.0039\text{ moles}}{250\text{ mL}}\times \frac{1000\text{ mL}}{1\text{ L}}=0.0156\text{ M}$

(c)

Interpretation Introduction

Interpretation : The mole of ${\text{Ca}}^{2+}$ in $25.00\text{ mL}$ should be determined during the reaction of $0.3946\text{ g}$ of sample of ${\text{CaCO}}_{\text{3}}$ with $12\text{ M HCl}$ .

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:

  $\text{Moles}=\frac{\text{mass}\left(\text{g}\right)}{\text{molar mass}\left(\text{g}/\text{mol}\right)}$

Answer to Problem 1ASA

In the given reaction mole of ${\text{Ca}}^{2+}$ in $25.00\text{ mL}$ should be $0.00039\text{ moles}$ .

Explanation of Solution

Given:

Molarity of ${\text{Ca}}^{2+}$ ions = $0.0156\text{ M}$

Volume = $25.00\text{ mL}$

Relation between molarity and volume is as follows:

  $\text{Moles}=\text{molarity}\times \text{volume}$

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

  $\begin{array}{c}\text{Moles}=\text{molarity}\times \text{volume}\\ =0.0156\frac{\text{mol}}{\text{L}}\times 0.025\text{ L}\\ \text{=0}\text{.00039 moles}\end{array}$