Chapter 12, Problem 75A

Interpretation Introduction

Interpretation : The days that will take to consume glucose are to be calculated.

Concept Introduction : The coefficients of a balanced chemical equation that are transformed into moles are used to calculate the mole ratio, which is a conversion factor.

When converting between a certain number of moles of one reactant or product and moles of another reactant or product, mole ratios are utilized.

The formula to calculate the number of moles is given as

  $\text{n}\text{=}\frac{\text{given}\text{mass}}{\text{molar mass}}$   ...... (1)

Answer to Problem 75A

The days that will take to consume glucose are $13\text{days}\text{8}\text{hr}$ .

Explanation of Solution

It takes $5\text{hr}$ to produce $8.0\text{kg}$ of alcohol.

The mass of glucose is $1.0\times {10}^3\text{kg}$ .

The chemical equation involving the production of ethyl alcohol is given as follows:

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6 }}\stackrel{\text{enzyme}}{\to }{\text{2C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH+2CO}}_{\text{2}}$

Molar mass of $\text{C}$ is $\text{12}\text{.00}\text{g/mol}$ .

Molar mass of $\text{H}$ is $\text{1}\text{.008}\text{g/mol}$ .

Molar mass of $\text{O}$ is $\text{16}\text{.00}\text{g/mol}$ .

Mass of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ is given as follows:

  $\begin{array}{c}\text{Mass of}{\text{C}}_6{\text{H}}_{12}{\text{O}}_6=6\text{M}\left(\text{C}\right)\text{+12M}\left(\text{H}\right)+6\text{M}\left(\text{O}\right)\\ =\left(\left(6\times 12.01\right)+\left(12\times 1.008\right)+\left(6\times 16.00\right)\right)\text{g/mol}\\ \text{= 180}\text{.156}\text{g/mol}\end{array}$

Mass of ${\text{C}}_2{\text{H}}_5\text{OH}$ is given as follows:

  $\begin{array}{c}\text{Mass of}{\text{C}}_2{\text{H}}_5\text{OH}=2\text{M}\left(\text{C}\right)\text{+6M}\left(\text{H}\right)+\text{M}\left(\text{O}\right)\\ =\left(\left(2\times 12.01\right)+\left(6\times 1.008\right)+16.00\right)\text{g/mol}\\ \text{= 46}\text{.068}\text{g/mol}\end{array}$

To calculate the number of moles of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$, substitute the values in equation (1):

  $\begin{array}{c}{\text{n}}_{{\text{C}}_6{\text{H}}_{12}{\text{O}}_6}\text{=}\frac{{\text{m}}_{{\text{C}}_6{\text{H}}_{12}{\text{O}}_6}}{\text{M}\left({\text{C}}_6{\text{H}}_{12}{\text{O}}_6\right)}\\ =\frac{1.0\times {10}^3\times {10}^3\cancel{\text{g}}}{180.156\cancel{\text{g}}\text{/mol}}\\ =5550.745\text{mol}\end{array}$

From the chemical equation, $\text{1}\text{mole}$ of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ gives $2\text{mole}$ of ${\text{C}}_2{\text{H}}_5\text{OH}$ .

So, the number of moles of ${\text{C}}_2{\text{H}}_5\text{OH}$ is given as

  $\begin{array}{c}{\text{n}}_{{\text{C}}_2{\text{H}}_5\text{OH}}=2\times {\text{n}}_{{\text{C}}_6{\text{H}}_{12}{\text{O}}_6}\\ =2\times 5550.745\text{mol}\\ =11101.490\text{mol}\end{array}$

To calculate the mass of ${\text{C}}_2{\text{H}}_5\text{OH}$, substitute the values in equation (1):

  $\begin{array}{c}{\text{m}}_{{\text{C}}_2{\text{H}}_5\text{OH}}={\text{n}}_{{\text{C}}_2{\text{H}}_5\text{OH}}\times \text{M}\left({\text{C}}_2{\text{H}}_5\text{OH}\right)\\ \text{=}11101.490\cancel{\text{mol}}\times \text{46}\text{.068}\text{g/}\cancel{\text{mol}}\\ =511.423\times {10}^3\text{g}\\ \text{= 511}\text{.423}\text{kg}\end{array}$

It takes $5\text{hr}$ to produce $8.0\text{kg}$ of alcohol.

For $\text{511}\text{.423}\text{kg}$ of alcohol, the time taken is given as

  $\begin{array}{c}\frac{5\text{hr}}{8.0\text{kg}}=\frac{x}{511.423\text{kg}}\\ x=\frac{5\text{hr}\times 511.423\cancel{\text{kg}}}{8.0\cancel{\text{kg}}}\\ =320\text{hr}\end{array}$

To calculate the time taken in days, divide the above value by 24.

  $\begin{array}{c}x=320\text{hr}\\ \text{=}320\cancel{\text{hr}}\times \frac{1\text{day}}{24\cancel{\text{hr}}}\left(\because 1\text{day}\text{=}\text{24}\text{hr}\right)\\ x=13\text{days}\text{8}\text{hr}\end{array}$