Chapter 8.4, Problem 8.7WE

Alka-Seltzer^® tablets contain aspirin, sodium bicarbonate, and citric acid. When they come into contact with water, the sodium bicarbonate (NaHCO₃) and citric acid (H₃C₆H₅O₇) react to form carbon dioxide gas, among other products.

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

The formation of CO₂ causes the trademark fizzing when the tablets are dropped into a glass of water. An Alka-Seltzer tablet contains 1.700 g of sodium bicarbonate and 1.000 g of citric acid. Determine, for a single tablet dissolved in water, (a) which ingredient is the limiting reactant, (b) what mass of the excess reactant is left over when the reaction is complete, and (c) what mass of CO₂ forms.

Interpretation Introduction

Interpretation:

For the given reaction, the limiting reactant, the number of moles of excess reactant and the mass of product ${\text{CO}}_{\text{2}}$ formed are needed to be determined.

Concept introduction:

• Limiting reactant: limiting reactant is a reactant, which consumes completely in the chemical reaction. The quantity of the product depends on this limiting reactant, it can be determined with the help of balanced chemical equation for the reaction.
• Excess reactant: is the reactant, which remains unreacted after the completion of a reaction.
• Balanced chemical equation of a reaction is written according to law of conservation of mass.
• Mole ratios between the reactants of a reaction are depends upon the coefficients of respective reactant in a balanced chemical equation.
• Equation for Number of moles of a substance, from its given mass is,

$\text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

To find: the limiting reactant for the given reaction.

Explanation of Solution

The mass of ${\text{NaHCO}}_{\text{3}}$ reacts in the reaction is given as $\text{1}\text{.7g}$.

The mass of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ reacts in the reaction is given as $\text{1g}$.

Number of moles of a substance, from its given mass is,

$\text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

Therefore,

The number of moles of ${\text{NaHCO}}_{\text{3}}$ is,

$\frac{\text{1}\text{.7}}{\text{84}\text{.01}}\text{=}\text{0}\text{.02024 mol}$

The number of moles of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ is,

$\frac{\text{1g}}{\text{192}\text{.12g}}\text{=}\text{0}\text{.005205 mol}$

To find: the limiting reactant for the given reaction.

The liming reactant of the reaction is ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$.

Explanation:

The balanced equation of the reaction is given as,

${\text{3NaHCO}}_{\text{3}}\text{+}{\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}\to {\text{3CO}}_{\text{2}}\text{+}{\text{3H}}_{\text{2}}\text{O}\text{+}{\text{Na}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$

The mole ratio between reactants ${\text{NaHCO}}_{\text{3}}$ and ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ is 3:1.

That is,

Three moles of ${\text{NaHCO}}_{\text{3}}$ molecules reacts with one mole of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$molecule to form the product molecules.

The number of moles of ${\text{NaHCO}}_{\text{3}}$ in the reaction is found as $\text{0}\text{.02024 mol}$.

The number of moles of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ in the reaction is found as $\text{0}\text{.005205 mol}$.

Therefore,

The number of moles of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ required to react with $\text{0}\text{.02024 mol}$ of ${\text{NaHCO}}_{\text{3}}$ to form products is,

$\frac{\text{1}}{\text{3}}\text{×}\text{0}\text{.02024 mol}\text{=}\text{0}\text{.006745}\text{mol}$

But there are only $\text{0}\text{.005205 mol}$ of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ molecules presented in the reaction mixture; thereby all the ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ ${\text{O}}_3$ molecules will be consumed completely during the reaction. So the limiting reactant of the reaction is ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$.

Hence,

The liming reactant of the reaction is ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$.

Interpretation Introduction

Interpretation:

For the given reaction, the limiting reactant, the number of moles of excess reactant and the mass of product ${\text{CO}}_{\text{2}}$ formed are needed to be determined.

Concept introduction:

• Limiting reactant: limiting reactant is a reactant, which consumes completely in the chemical reaction. The quantity of the product depends on this limiting reactant, it can be determined with the help of balanced chemical equation for the reaction.
• Excess reactant: is the reactant, which remains unreacted after the completion of a reaction.
• Balanced chemical equation of a reaction is written according to law of conservation of mass.
• Mole ratios between the reactants of a reaction are depends upon the coefficients of respective reactant in a balanced chemical equation.
• Equation for Number of moles of a substance, from its given mass is,

$\text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

To determine: the mass of excess reactant in the given reaction.

Explanation of Solution

The balanced equation of the reaction is given as,

${\text{3NaHCO}}_{\text{3}}\text{+}{\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}\to {\text{3CO}}_{\text{2}}\text{+}{\text{3H}}_{\text{2}}\text{O}\text{+}{\text{Na}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$

The mole ratio between reactants ${\text{NaHCO}}_{\text{3}}$ and ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ is 3:1.

That is,

Three moles of ${\text{NaHCO}}_{\text{3}}$ molecules reacts with one mole of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$molecule to form the product molecules.

The liming reactant of the reaction is found as ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$.

The number of moles of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ in the reaction is found as $\text{0}\text{.005205 mol}$.

The number of moles of ${\text{NaHCO}}_{\text{3}}$ in the reaction is found as $\text{0}\text{.02024 mol}$.

Therefore,

The number of mole of ${\text{NaHCO}}_{\text{3}}$ reacted with $\text{0}\text{.005205 mol}$ of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ is,

$\text{3}\text{×}\text{0}\text{.005205 mol}\text{=}\text{0}\text{.01562}\text{mol}$

The number of moles of ${\text{NaHCO}}_{\text{3}}$ in the reaction mixture is $\text{0}\text{.02024 mol}$.

So, the number of moles unreacted or the excess reactant ${\text{NaHCO}}_{\text{3}}$ remaining at the end of reaction is,

$\text{0}\text{.02024}\text{mol}\text{-}\text{0}\text{.01562}\text{mol}\text{=}\text{0}\text{.00462}\text{mol}$

Equation for Number of grams of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{Number}\text{of}\text{grams}$

Then,

The mass of ${\text{NaHCO}}_{\text{3}}$ is,

$\text{0}\text{.00462}\text{×}\text{84}\text{.01}\text{g}\text{=}\text{0}\text{.0388g}$

Interpretation Introduction

Interpretation:

For the given reaction, the limiting reactant, the number of moles of excess reactant and the mass of product ${\text{CO}}_{\text{2}}$ formed are needed to be determined.

Concept introduction:

• Limiting reactant: limiting reactant is a reactant, which consumes completely in the chemical reaction. The quantity of the product depends on this limiting reactant, it can be determined with the help of balanced chemical equation for the reaction.
• Excess reactant: is the reactant, which remains unreacted after the completion of a reaction.
• Balanced chemical equation of a reaction is written according to law of conservation of mass.
• Mole ratios between the reactants of a reaction are depends upon the coefficients of respective reactant in a balanced chemical equation.
• Equation for Number of moles of a substance, from its given mass is,

$\text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

To determine: the mass of product ${\text{CO}}_{\text{2}}$ formed in the given reaction.

The mass of product ${\text{CO}}_{\text{2}}$ formed is 0.6874 g

Explanation of Solution

The balanced equation of the reaction is given as,

${\text{3NaHCO}}_{\text{3}}\text{+}{\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}\to {\text{3CO}}_{\text{2}}\text{+}{\text{3H}}_{\text{2}}\text{O}\text{+}{\text{Na}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$

The mole ratio between reactant ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ and product ${\text{CO}}_{\text{2}}$ is 1:3.

That is,

One mole of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$molecule reactswith ${\text{NaHCO}}_{\text{3}}$ to form the product1 mole of ${\text{CO}}_{\text{2}}$ and other molecules.

The liming reactant of the reaction is found as ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$.

The number of moles of ${\text{H}}_{\text{3}}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{7}}$ in the reaction is found as $\text{0}\text{.005205 mol}$.

Therefore,

The number of moles of product ${\text{CO}}_{\text{2}}$ formed is,

$3\text{×}\text{0}\text{.005205 mol}\text{=}\text{0}\text{.01562}\text{mol}$

Equation for Number of grams of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{Number}\text{of}\text{grams}$

Then,

The mass of ${\text{CO}}_{\text{2}}$ is,

$\text{0}\text{.01562}\text{×}\text{44}\text{.01}\text{g}\text{=}\text{0}\text{.6874g}$