Chapter 11.3, Problem 24PP

(a)

Interpretation Introduction

Interpretation:

The balanced chemical equation for given reaction is to be stated.

Concept introduction:

A balanced chemical equation consists of equal number of atoms of all the elements that are present in a chemical reaction.

Answer to Problem 24PP

The balanced chemical equation for given reaction is shown below:

${\text{6CO}}_{\text{2}}\left(\text{g}\right)+{\text{6H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{aq}\right)+{\text{6O}}_2\left(\text{g}\right)$

Explanation of Solution

The reactants are carbon dioxide and water and the reaction leads to the formation of glucose and oxygen. The chemical equation corresponding to given reaction is shown below:

${\text{6CO}}_{\text{2}}\left(\text{g}\right)+{\text{6H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{aq}\right)+{\text{6O}}_2\left(\text{g}\right)$

(b)

Interpretation Introduction

Interpretation:

The limiting reagent in given reaction is to be determined.

Concept introduction:

The formula to calculate the number of moles is given as shown below:

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

Answer to Problem 24PP

Carbon dioxide acts as the limiting reagent.

Explanation of Solution

The molar mass of carbon dioxide is $\text{44}\text{.0}\text{g/mol}$. Substitute the given mass and molar mass in the above equation for number of moles.

$\begin{array}{c}\text{Number of moles}=\frac{\text{88}\text{.0}\text{g}}{44.0\frac{\text{g}}{\text{mol}}}\\ =2\text{mol}\end{array}$

The molar mass of water is $\text{18}\text{.0}\text{g/mol}$. Substitute the given mass and molar mass in the above equation for number of moles.

$\begin{array}{c}\text{Number of moles}=\frac{\text{64}\text{.0}\text{g}}{18.0\frac{\text{g}}{\text{mol}}}\\ =3.56\text{mol}\end{array}$

The limiting reagent is determined as shown below:

$\begin{array}{c}6\text{mol}{\text{CO}}_{\text{2}}=6\text{mol}{\text{H}}_{\text{2}}\text{O}\\ \text{2}\text{.0}\text{mol}{\text{CO}}_{\text{2}}=\frac{6}{6}\times 2\text{mol}{\text{H}}_{\text{2}}\text{O}\\ =2\text{mol}{\text{H}}_{\text{2}}\text{O}\end{array}$

Thus, the given amount of water is more than required. Therefore, carbon dioxide acts as the limiting reagent.

(c)

Interpretation Introduction

Interpretation:

The excess reagent in given reaction is to be determined.

Concept introduction:

The reagent that gets completely used in reaction is the limiting reagent. The reagent that is left even after the reaction is defined as the excess reagent.

Answer to Problem 24PP

Water acts as the excess reagent.

Explanation of Solution

The given moles of water are $3.56\text{mol}$ and the required number of moles is two. Thus, $1.56\text{mol}$ of water are excess. Therefore, water acts as the excess reagent.

(d)

Interpretation Introduction

Interpretation:

The mass of excess reagent is to be stated.

Concept introduction:

The reagent that gets completely used in reaction is the limiting reagent. The reagent that is left even after the reaction is defined as the excess reagent.

Answer to Problem 24PP

The mass of excess reagent is $28.08\text{g}$.

Explanation of Solution

The given moles of water are $3.56\text{mol}$ and the required number of moles is two. Thus, $1.56\text{mol}$ of water are excess. Therefore, water acts as the excess reagent. The molar mass of water is $\text{18}\text{.0}\text{g/mol}$. Substitute the number of moles and molar mass in the above equation for number of moles.

$\begin{array}{c}\text{1}\text{.56}=\frac{\text{Given mass}}{18.0\frac{\text{g}}{\text{mol}}}\\ \text{Given mass}=28.08\text{g}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The mass of glucose formed is to be stated.

Concept introduction:

The reagent that gets completely used in reaction is the limiting reagent. The reagent that is left even after the reaction is defined as the excess reagent.

Answer to Problem 24PP

The mass of glucose formed is $59.45\text{g}$.

Explanation of Solution

The moles of glucose formed are calculated as shown below:

$\begin{array}{c}6\text{mol}{\text{CO}}_{\text{2}}=1\text{mol}\text{glucose}\\ \text{2}\text{.0}\text{mol}{\text{CO}}_{\text{2}}=\frac{1}{6}\times 2\text{mol}\text{glucose}\\ =0.33\text{mol}\text{glucose}\end{array}$

The molar mass of glucose is $\text{180}\text{.156}\text{g/mol}$. Substitute the number of moles and molar mass in the above equation for number of moles.

$\begin{array}{c}\text{0}\text{.33}\text{mol}=\frac{\text{Given mass}}{180.156\frac{\text{g}}{\text{mol}}}\\ \text{Given mass}=59.45\text{g}\end{array}$