Chapter 4, Problem 86GQ

(a)

Interpretation Introduction

Interpretation:

The reactants involved in the reaction should be named.

Answer to Problem 86GQ

Sodium iodate ($NaI{O}_3$) and sodium bisulfite ($NaHS{O}_3$ ) are the reactants involved in the reaction.

Explanation of Solution

Balanced chemical equation for the given reaction is,

  $2NaI{O}_{3(aq)}+5NaHS{O}_{3(aq)}\stackrel{}{\to }3NaHS{O}_{4(aq)}+2N{a}_{2}S{O}_{4(aq)}+H_2{O}_{(l)}+I_{2(aq)}$

Reactant of a chemical reaction is the substrate compounds or the compounds which undergo a chemical reaction.

Here the reactants involved in this reaction are Sodium iodate ($NaI{O}_3$) and sodium bisulfite ($NaHS{O}_3$ ) respectively.

(b)

Interpretation Introduction

Interpretation:

The masses of $NaI{O}_3$ and $NaHS{O}_3$ required to produce $1kgof{I}_2$ has to be calculated.

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.
• 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

Answer to Problem 86GQ

Mass of $NaI{O}_3$ and $NaHS{O}_3$ required to produce $1kg$ of ${\text{I}}_{\text{2}}$ are $\text{1}\text{.55941}k\text{g}$ and $\text{2}\text{.05005}k\text{g}$ respectively.

Explanation of Solution

Balanced chemical equation for the given reaction is,

    $2NaI{O}_{3(aq)}+5NaHS{O}_{3(aq)}\stackrel{}{\to }3NaHS{O}_{4(aq)}+2N{a}_{2}S{O}_{4(aq)}+H_2{O}_{(l)}+I_{2(aq)}$

From the balanced chemical equation it is clear that $I_2$ reacts with $NaI{O}_3$ and $NaHS{O}_3$ in $\text{1:2}\text{ratio}$ and $1:5ratio$ respectively.

The mass of $NaI{O}_3$ required to produce $1kgof{I}_2$ is,

  $\begin{array}{l}\left(\frac{1000g{I}_2}{253.8g/mol{I}_2}\right)\times \left(\frac{2molNaI{O}_3}{1mol{I}_2}\right)\times \left(\frac{197.89gNaI{O}_3}{1molNaI{O}_3}\right)=1559.41g\\ \text{=1}\text{.55941}k\text{g}{\text{NaIO}}_{\text{3}}\end{array}$

The mass of $NaHS{O}_3$ required to produce $1kgof{I}_2$ is,

  $\begin{array}{l}\left(\frac{1000g{I}_2}{253.8g/mol{I}_2}\right)\times \left(\frac{5molNaHS{O}_3}{1mol{I}_2}\right)\times \left(\frac{104.061gNaHS{O}_3}{1molNaI{O}_3}\right)=2050.05g\\ \text{=}\text{2}\text{.05005}k\text{g}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The theoretical yield of $I_2$ produced in the reaction with $15.0gofNaI{O}_3$ and $125mLof0.853M$ has to be determined.

Concept Introduction:

Theoretical yield is the maximum product yield that can be expected based on the masses of the reactants and the reaction stoichiometry.

Answer to Problem 86GQ

Theoretical yield of $I_2$ produced in the reaction with $15.0gofNaI{O}_3$ and $125mLof0.853M$ is $\text{5}\text{.41}\text{g}$

Explanation of Solution

Balanced chemical equation for the given reaction is,

  $2NaI{O}_{3(aq)}+5NaHS{O}_{3(aq)}\stackrel{}{\to }3NaHS{O}_{4(aq)}+2N{a}_{2}S{O}_{4(aq)}+H_2{O}_{(l)}+I_{2(aq)}$

The amount (moles) of $NaI{O}_3$ used in the reaction can be calculated as follows,

$\begin{array}{l}AmountofNaI{O}_{3}usedinthereaction=\frac{15.0gNaI{O}_3}{197.89g/molNaI{O}_3}\\ =0.0757molNaI{O}_3\end{array}$

Amount of $NaHS{O}_3$ can be calculated from its volume and concentration as follows,

  $\begin{array}{l}\text{Amount}\text{of}{\text{NaHSO}}_{\text{3}}\text{=}{\text{C}}_{{\text{NaHSO}}_{\text{3}}}\text{×}{\text{V}}_{{\text{NaHSO}}_{\text{3}}}\\ \text{=}\frac{\text{0}\text{.853}\text{mol}}{\cancel{\text{L}}}\text{×}\text{0}\text{.125}\cancel{\text{L}}\\ \text{=}\text{0}\text{.1066}\text{mol}{\text{NaHSO}}_{\text{3}}\end{array}$

$\text{0}\text{.1066}\text{mol}{\text{NaHSO}}_{\text{3}}$ would react completely with $\text{0}\text{.04264}\text{mol}{\text{NaIO}}_{\text{3}}$

  $\text{0}\text{.1066}\text{mol}{\text{NaHSO}}_{\text{3}}\text{×}\frac{\text{2}\text{mol}{\text{NaIO}}_{\text{3}}}{\text{5}\text{mol}{\text{NaHSO}}_{\text{3}}}=\text{0}\text{.04264}\text{mol}{\text{NaIO}}_{\text{3}}$

Here more $NaI{O}_3$ present, therefore $NaHS{O}_3$ is the limiting reagent.

Using the stoichiometric factor the amount of $I_2$ can be calculated as follows,

  $\text{0}\text{.1066}\text{mol}{\text{NaHSO}}_{\text{3}}\text{×}\frac{\text{1}\text{mol}{\text{I}}_{\text{2}}}{\text{5}\text{mol}{\text{NaHSO}}_{\text{3}}}\text{=}\text{0}\text{.02132}\text{mol}{\text{I}}_{\text{2}}$

The mass of $I_2$ produced in the reaction is $\text{5}\text{.41}\text{g}$.  It is obtained by multiplying the amount of $I_2$ with its molar mass (molar mass of $I_2$ is $\text{253}\text{.80}\text{g/mol}$ )

Therefore the theoretical yield of $I_2$ produced in the reaction with $15.0gofNaI{O}_3$ and $125mLof0.853M$ is $\text{5}\text{.41}\text{g}$