Question

The fermentation of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) produces ethyl alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and \(\mathrm{CO}_{2}\) : $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g) $$ (a) How many moles of \(\mathrm{CO}_{2}\) are produced when \(0.400\) mol of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) reacts in this fashion? (b) How many grams of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) are needed to form \(7.50 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) ? (c) How many grams of \(\mathrm{CO}_{2}\) form when \(7.50 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) are produced?

Short answer

(a) 0.800 moles of CO2 are produced when 0.400 mol of C6H12O6 reacts. (b) 14.66 grams of C6H12O6 are needed to form 7.50 g of C2H5OH. (c) 7.16 grams of CO2 are formed when 7.50 g of C2H5OH is produced.

Step-by-step solution

a) The moles of CO2 produced when 0.400 mol of C6H12O6 reacts

Using the stoichiometry from the balanced chemical equation, we can find the moles of CO2 produced when 0.400 mol of C6H12O6 reacts. According to the equation, one mole of C6H12O6 produces two moles of CO2: \[ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g) \] So, to find the moles of CO2 produced, we will set up a conversion factor using the stoichiometry: \[ 0.400 \,\text{mol}\, C_{6} H_{12} O_{6} \times \frac{2\,\text{mol}\,CO_{2}}{1\,\text{mol}\,C_{6} H_{12} O_{6}} \] This simplifies to: \[ 0.400 \,\text{mol}\, C_{6} H_{12} O_{6} \times 2 = 0.800\,\text{mol}\,CO_{2} \] So, 0.800 moles of CO2 are produced when 0.400 mol of C6H12O6 reacts.

b) The grams of C6H12O6 needed to form 7.50 g of C2H5OH

First, we need to convert 7.50 g of C2H5OH to moles. To do this, we use the molar mass of C2H5OH, which is approximately 46.07 g/mol: \[ \frac{7.50\,\text{g}\,C_{2} H_{5} OH}{46.07\,\text{g/mol}} = 0.1628\,\text{mol}\,C_{2} H_{5} OH \] Now, using the stoichiometry from the balanced chemical equation, we can find out the moles of C6H12O6 needed to form 0.1628 mol of C2H5OH: \[ \frac{1\,\text{mol}\,C_{6} H_{12} O_{6}}{2\,\text{mol}\,C_{2} H_{5} OH} \times 0.1628\,\text{mol}\,C_{2} H_{5} OH \] This simplifies to: \[ \frac{1}{2} \times 0.1628\,\text{mol}\,C_{2} H_{5} OH = 0.0814\,\text{mol}\,C_{6} H_{12} O_{6} \] Finally, we convert 0.0814 mol of C6H12O6 to grams, using its molar mass, which is approximately 180.2 g/mol: \[ 0.0814\,\text{mol}\,C_{6} H_{12} O_{6} \times 180.2\,\text{g/mol} = 14.66\,\text{g}\,C_{6} H_{12} O_{6} \] So, 14.66 grams of C6H12O6 are needed to form 7.50 g of C2H5OH.

c) The grams of CO2 formed when 7.50 g of C2H5OH is produced

We have already calculated the moles of C2H5OH produced, which is 0.1628 mol. Using the stoichiometry from the balanced chemical equation, we can find out the moles of CO2 produced along with 0.1628 mol of C2H5OH: \[ \frac{2\,\text{mol}\,CO_{2}}{2\,\text{mol}\,C_{2} H_{5} OH} \times 0.1628\,\text{mol}\,C_{2} H_{5} OH \] This simplifies to: \[ 1 \times 0.1628\,\text{mol}\,C_{2} H_{5} OH = 0.1628\,\text{mol}\,CO_{2} \] Now, we convert 0.1628 mol of CO2 to grams, using its molar mass, which is approximately 44.01 g/mol: \[ 0.1628\,\text{mol}\,CO_{2} \times 44.01\,\text{g/mol} = 7.16\,\text{g}\,CO_{2} \] So, 7.16 grams of CO2 are formed when 7.50 g of C2H5OH is produced.

Key concepts

Mole Concept

The mole concept is a fundamental concept in chemistry that gives a bridge between the atomic world, visible only at microscopic levels, and the macroscopic world that we can actually measure. One mole represents a quantity of \(6.022 \times 10^{23}\) entities, be it atoms, molecules, or other basic units. This number is known as Avogadro's number.

To understand chemical reactions, the mole concept helps us in comparing products and reactants in a chemical equation. For instance, in the fermentation reaction of glucose \(C_{6} H_{12} O_{6} \rightarrow 2C_{2} H_{5} OH + 2CO_{2}\), every mole of glucose can produce two moles of ethyl alcohol (ethanol) and two moles of carbon dioxide.

Here’s how it works practically:

• For every 1 mole of \(C_{6} H_{12} O_{6}\), 2 moles of \(CO_{2}\) are produced.
• It enables us to use mass, the more tangible measurement, to quantify chemical reactions.

The use of the mole allows chemists to count particles by weighing them, simplifying the task of calculating how much product is formed from a certain amount of reactant.

Molar Mass Calculations

Molar mass is the mass of one mole of a given substance and is usually expressed in grams per mole (g/mol). This measurement acts as the bridge between the number of moles and the mass of a material.

Let's take ethanol (\(C_{2} H_{5} OH\)) as an example for molar mass calculation. To find its molar mass, you need to add up the atomic masses of all the atoms present in the molecule:

• Carbon (C): 2 atoms \( \times \; 12.01 \; \text{g/mol} = 24.02 \; \text{g/mol}\)
• Hydrogen (H): 6 atoms \( \times \; 1.01 \; \text{g/mol} = 6.06 \; \text{g/mol}\)
• Oxygen (O): 1 atom (\( \times \; 16.00 \; \text{g/mol} = 16.00 \; \text{g/mol}\)

Total Molar Mass of \(C_{2} H_{5} OH\) = 24.02 g/mol + 6.06 g/mol + 16.00 g/mol = 46.08 g/mol.

This information allows us to convert grams to moles and vice versa. For example, in the given reaction scenario, when dealing with the grams of ethanol, you can convert this to moles: if 7.50 g of \(C_{2} H_{5} OH\) is given, convert it using its molar mass to find moles.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They show the reactants and products involved in the reaction and their stoichiometric relationships. Balanced chemical equations maintain the principle of conservation of mass, ensuring that the number of atoms for each element is the same on both sides of the equation.

In the case of the fermentation of glucose: \(C_{6} H_{12} O_{6} (aq) \rightarrow 2 C_{2} H_{5} OH (aq) + 2 CO_{2} (g)\), the equation reflects the stochiometric balance:

• One mole of glucose decomposes into two moles of ethyl alcohol and two moles of carbon dioxide.

Balancing chemical equations involves making sure the number of atoms for every element is the same in reactants and products. This is important for correct stoichiometric calculations, allowing you to convert between moles, mass, and volume using balanced relations.

Balancing chemical equations is crucial because it enables chemists to follow the law of conservation of mass and accurately describe the flow of reactants to form products in a chemical reaction.