Chapter 10, Problem 74E

Interpretation Introduction

(a)

Interpretation:

The mass of carbon dioxide that would produce when $6.00\text{g}$ of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ reacts with $20.0\text{g}$ of ${\text{NaHCO}}_{\text{3}}$ is to be calculated.

Concept introduction:

The limiting reagent of a reaction is that reactant of the reaction that controls the amount of product formed. The limiting agents limit the amount of product and by adding some more amount of the limiting reagent in the reaction mixture, the amount of product can be increased.

Answer to Problem 74E

The mass of carbon dioxide that would produce when $6.00\text{g}$ of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ reacts with $20.0\text{g}$ of ${\text{NaHCO}}_{\text{3}}$ is $4.12\text{g}$.

Explanation of Solution

The reaction between ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ and ${\text{NaHCO}}_{\text{3}}$ is shown below.

${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}\left(\text{aq}\right)+3{\text{NaHCO}}_{\text{3}}\left(\text{aq}\right)\to {\text{Na}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}\left(\text{aq}\right)+3{\text{CO}}_{\text{2}}\left(\text{g}\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The mass of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ reacted is $6.00\text{g}$.

The mass of ${\text{NaHCO}}_{\text{3}}$ reacted is $20.0\text{g}$.

The molar mass of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ is $192.12\text{g}/\text{mol}$.

The molar mass of ${\text{NaHCO}}_{\text{3}}$ is $84.01\text{g}/\text{mol}$.

The molar mass of carbon dioxide is $44.01\text{g}/\text{mol}$.

The number of moles of a substance is given by the expression shown below.

$\text{n}=\frac{\text{m}}{\text{M}}$…(1)

Where,

• $\text{m}$ is the mass of the substance.

• $\text{M}$ is the molar mass of the substance.

Substitute the mass and molar mass of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ in the equation (1).

$\begin{array}{rll}\text{n}& =& \frac{6.00\text{g}}{192.12\text{g}/\text{mol}}\\ & =& 0.0312\text{mol}\end{array}$

Therefore, the number of moles of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ present in the reaction mixture is $0.0312\text{mol}$.

Substitute the mass and molar mass of ${\text{NaHCO}}_{\text{3}}$ in the equation (1).

$\begin{array}{rll}\text{n}& =& \frac{20.0\text{g}}{84.01\text{g}/\text{mol}}\\ & =& 0.2381\text{mol}\end{array}$

Therefore, the number of moles of ${\text{NaHCO}}_{\text{3}}$ present in the reaction mixture is $0.2381\text{mol}$.

One mole of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ reacts with three moles of ${\text{NaHCO}}_{\text{3}}$. Therefore, the relation between the number of moles of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ and ${\text{NaHCO}}_{\text{3}}$ is given by the expression shown below.

${\text{n}}_{{\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}}=\frac{{\text{n}}_{{\text{NaHCO}}_{\text{3}}}}{3}$…(2)

Where,

• ${\text{n}}_{{\text{NaHCO}\text{3}{}_{}}_{}}$ is the number of moles of ${\text{NaHCO}}_{\text{3}}$.

• ${\text{n}}_{{\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}}$ is the number of moles of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$.

Substitute the value of ${\text{n}}_{{\text{NaHCO}}_{\text{3}}}$ in the equation (2).

$\begin{array}{rll}{\text{n}}_{{\text{CO}}_{\text{2}}}& =& \frac{0.2381\text{mol}}{3}\\ & =& 0.0794\text{mol}\end{array}$

The required amount of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ is greater than the available amount of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$. Therefore, ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ is the limiting reagent.

Three moles of carbon dioxide is produced by one mole of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$. Therefore, the relation between the number of moles of carbon dioxide and ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ is given by the expression shown below.

${\text{n}}_{{\text{CO}}_{\text{2}}}=3{\text{n}}_{{\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}}$…(3)

Where,

• ${\text{n}}_{{\text{CO}\text{2}{}_{}}_{}}$ is the number of moles of carbon dioxide.

• ${\text{n}}_{{\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}}$ is the number of moles of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$.

Substitute the value of ${\text{n}}_{{\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}}$ in the equation (3).

$\begin{array}{rll}{\text{n}}_{{\text{CO}}_{\text{2}}}& =& \left(3\right)\left(0.0312\text{mol}\right)\\ & =& 0.0936\text{mol}\end{array}$

Rearrange the equation (1) for the value of $\text{m}$.

$\text{m}=\text{nM}$…(4)

Substitute the value of molar mass and the number of moles of carbon dioxide in the equation (4).

$\begin{array}{rll}\text{m}& =& \left(0.0936\text{mol}\right)\left(44.01\text{g}/\text{mol}\right)\\ & =& 4.12\text{g}\end{array}$

Therefore, the mass of carbon dioxide that would produce when $6.00\text{g}$ of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ reacts with $20.0\text{g}$ of ${\text{NaHCO}}_{\text{3}}$ is $4.12\text{g}$.

Conclusion

The mass of carbon dioxide that would produce when $6.00\text{g}$ of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ reacts with $20.0\text{g}$ of ${\text{NaHCO}}_{\text{3}}$ is $4.12\text{g}$.

Interpretation Introduction

(b)

Interpretation:

The mass of reactant which would remain unreacted in the reaction of $6.00\text{g}$ of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ and $20.0\text{g}$ of ${\text{NaHCO}}_{\text{3}}$ is to be stated.

Concept introduction:

The limiting reagent of a reaction is that reactant of the reaction that controls the amount of product formed. The limiting agents limit the amount of product and by adding some more amount of the limiting reagent in the reaction mixture, the amount of product can be increased.

Answer to Problem 74E

The mass of ${\text{NaHCO}}_{\text{3}}$ present in excess is $12.1367\text{g}$.

Explanation of Solution

The reaction between ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ and ${\text{NaHCO}}_{\text{3}}$ is shown below.

${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}\left(\text{aq}\right)+3{\text{NaHCO}}_{\text{3}}\left(\text{aq}\right)\to {\text{Na}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}\left(\text{aq}\right)+3{\text{CO}}_{\text{2}}\left(\text{g}\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The mass of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ reacted is $6.00\text{g}$.

The mass of ${\text{NaHCO}}_{\text{3}}$ reacted is $20.0\text{g}$.

The molar mass of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ is $192.12\text{g}/\text{mol}$.

The molar mass of ${\text{NaHCO}}_{\text{3}}$ is $84.01\text{g}/\text{mol}$.

The molar mass of carbon dioxide is $44.01\text{g}/\text{mol}$.

The number of moles of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ present in the reaction mixture is $0.0312\text{mol}$.

One mole of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ reacts with three moles of ${\text{NaHCO}}_{\text{3}}$. Therefore, the relation between the number of moles of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ and ${\text{NaHCO}}_{\text{3}}$ is given by the expression shown below.

${\text{n}}_{{\text{NaHCO}}_{\text{3}}}=3{\text{n}}_{{\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}}$

Where,

• ${\text{n}}_{{\text{NaHCO}\text{3}{}_{}}_{}}$ is the number of moles of ${\text{NaHCO}}_{\text{3}}$.

• ${\text{n}}_{{\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}}$ is the number of moles of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$.

Substitute the value of ${\text{n}}_{{\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}}$ in the above equation.

$\begin{array}{rll}{\text{n}}_{{\text{NaHCO}}_{\text{3}}}& =& \left(3\right)\left(0.0312\text{mol}\right)\\ & =& 0.0936\text{mol}\end{array}$

The number of moles of ${\text{NaHCO}}_{\text{3}}$ reacted is $0.0936\text{mol}$.

The number of moles of a substance is given by the expression shown below.

$\text{n}=\frac{\text{m}}{\text{M}}$…(1)

Where,

• $\text{m}$ is the mass of the substance.

• $\text{M}$ is the molar mass of the substance.

Rearrange the equation (1) for the value of $\text{m}$.

$\text{m}=\text{nM}$…(4)

Substitute the value of molar mass and the number of moles of ${\text{NaHCO}}_{\text{3}}$ in the equation (4).

$\begin{array}{rll}\text{m}& =& \left(0.0936\text{mol}\right)\left(84.01\text{g}/\text{mol}\right)\\ & =& 7.8633\text{g}\end{array}$

Therefore, the mass of ${\text{NaHCO}}_{\text{3}}$ reacted to produce $4.12\text{g}$ of carbon dioxide is $7.8633\text{g}$.

The excess mass of ${\text{NaHCO}}_{\text{3}}$ is given by the expression as shown below.

${\text{m}}_{{\text{NaHCO}}_{\text{3}}}={\text{m}}_{\text{1}}-{\text{m}}_{\text{2}}$

Where,

• ${\text{m}}_{\text{1}}$ is the initial mass of ${\text{NaHCO}}_{\text{3}}$ present in the reaction mixture.

• ${\text{m}}_{\text{2}}$ is the mass of ${\text{NaHCO}}_{\text{3}}$ reacted in the reaction.

Substitute the value of ${\text{m}}_{\text{1}}$ and ${\text{m}}_{\text{2}}$ in the above equation.

$\begin{array}{rll}{\text{m}}_{{\text{NaHCO}}_{\text{3}}}& =& 20.0\text{g}-7.8633\text{g}\\ & =& 12.1367\text{g}\end{array}$

Therefore, the mass of ${\text{NaHCO}}_{\text{3}}$ present in excess is $12.1367\text{g}$.

Conclusion

The mass of ${\text{NaHCO}}_{\text{3}}$ present in excess is $12.1367\text{g}$.

Interpretation Introduction

(c)

Interpretation:

Whether the name of ${\text{Na}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ can be predicted or not is to be stated.

Concept introduction:

Ionic compounds are those compounds that have the strong electrostatic force of attraction between the oppositely charged ions. Ionic compounds conduct electricity in solution Salts are the ionic compounds that contain an anion part and cation part. Ionic compounds get dissociated into ions when they dissolved in a solution.

Answer to Problem 74E

The name of ${\text{Na}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ can be predicted. The name of ${\text{Na}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ is sodium citrate.

Explanation of Solution

The compound ${\text{Na}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ is salt.

The cation of ${\text{Na}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ is sodium ion $\left({\text{Na}}^{+}\right)$.

The anion of ${\text{Na}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ is the ion formed by citric acid. The name of the ion formed by citric acid is citrated ion $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}{}^{3-}\right)$.

Therefore, the name of ${\text{Na}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ is sodium citrate.

Conclusion

The name of ${\text{Na}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ can be predicted. The name of ${\text{Na}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ is sodium citrate.