Problem 24 - Chemistry Matter and Change

Question

Challenge Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) and oxygen. A plant has 88.0 \(\mathrm{g}\) of carbon dioxide and 64.0 \(\mathrm{g}\) of water available for photosynthesis. \begin{equation} \begin{array}{l}{\text { a. Write the balanced chemical equation for the reaction. }} \\ {\text { b. Determine the limiting reactant. }} \\ {\text { c. Determine the excess reactant. }} \\ {\text { d. Determine the mass in excess. }} \\ {\text { e. Determine the mass of glucose produced. }}\end{array} \end{equation}

Step-by-Step Solution

Answer

The limiting reactant is CO2, with an excess of 27.9 g of water, producing 59.9 g of glucose.

Step 1: Write Balanced Equation

The balanced chemical equation for photosynthesis is: \[ 6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \] This equation shows that 6 molecules of carbon dioxide react with 6 molecules of water to produce 1 molecule of glucose and 6 molecules of oxygen.

Step 2: Calculate Moles of Reactants

First, find the moles of carbon dioxide and water using their molar masses. The molar mass of \(\text{CO}_2\) is 44.01 g/mol, so the moles of \(\text{CO}_2\) are: \[ \text{Moles of CO}_2 = \frac{88.0\text{ g}}{44.01\text{ g/mol}} \approx 2.00\text{ mol} \] The molar mass of \(\text{H}_2\text{O}\) is 18.02 g/mol, so the moles of \(\text{H}_2\text{O}\) are: \[ \text{Moles of H}_2\text{O} = \frac{64.0\text{ g}}{18.02\text{ g/mol}} \approx 3.55\text{ mol} \]

Step 3: Determine Limiting Reactant

Using the stoichiometric coefficients from the balanced equation, compare the mole ratio. - The reaction requires 6 moles of \(\text{H}_2\text{O}\) for every 6 moles of \(\text{CO}_2\), which simplifies to a 1:1 ratio. - Since we have 2.00 moles of \(\text{CO}_2\) and 3.55 moles of \(\text{H}_2\text{O}\), and since 2.00 moles of \(\text{H}_2\text{O}\) are needed for 2.00 moles of \(\text{CO}_2\), \(\text{CO}_2\) is the limiting reactant.

Step 4: Determine Excess Reactant

The excess reactant is \(\text{H}_2\text{O}\) because more moles are available than required by the stoichiometry of the reaction. Based on the previous determination of the limiting reactant, there is more than enough \(\text{H}_2\text{O}\) available to react with the \(\text{CO}_2\).

Step 5: Calculate Mass in Excess

Determine how much \(\text{H}_2\text{O}\) is used: - Needed \(\text{H}_2\text{O}\) for 2.00 moles of \(\text{CO}_2\): 2.00 moles - Unused \(\text{H}_2\text{O}\) is \(3.55 - 2.00 = 1.55\) moles.Calculate the mass of excess \(\text{H}_2\text{O}\): \[ \text{Mass of excess } \text{H}_2\text{O} = 1.55 \text{ mol} \times 18.02 \text{ g/mol} \approx 27.9 \text{ g} \]

Step 6: Calculate Mass of Glucose Produced

Using the balanced equation, 1 mole of \(\text{C}_6\text{H}_{12}\text{O}_6\) is produced per 6 moles of \(\text{CO}_2\). Thus, for 2.00 moles of \(\text{CO}_2\), - \(\text{Moles of glucose} = \frac{2.00}{6} = 0.333\) moles - The molar mass of glucose is: \(180.18 \text{ g/mol}\)- \(\text{Mass of glucose} = 0.333 \times 180.18 \approx 59.9 \text{ g} \)

Key Concepts

Balanced Chemical Equation

In chemical reactions, a balanced chemical equation is crucial. It ensures the Law of Conservation of Mass is respected. This law states that matter cannot be created or destroyed.

The balanced equation for photosynthesis is:

• \[ 6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \]

Here, 6 molecules of carbon dioxide (\(\text{CO}_2\)) react with 6 water molecules (\(\text{H}_2\text{O}\)) to form one glucose molecule (\(\text{C}_6\text{H}_{12}\text{O}_6\)) and six oxygen molecules (\(\text{O}_2\)). This balance ensures equal numbers of each type of atom on both sides of the equation.

Balanced equations are not just numbers; they tell us "how much" of each ingredient you need to get the desired product, much like a recipe.

Limiting Reactant

The limiting reactant in a chemical reaction is the substance that is completely consumed first, stopping the reaction from continuing.

To find the limiting reactant, compare the mole ratio needed by the balanced equation with the actual moles you have. From the balanced equation:

• Every 6 moles of \(\text{CO}_2\) require 6 moles of \(\text{H}_2\text{O}\).

Given 2.00 moles of \(\text{CO}_2\) and 3.55 moles of \(\text{H}_2\text{O}\), \(\text{CO}_2\) runs out first since it no longer has a sufficient amount to react with all \(\text{H}_2\text{O}\). This makes \(\text{CO}_2\) the limiting reactant.

Identifying the limiting reactant is vital because it determines the maximum amount of product formed. Without it, you might make incorrect assumptions about how much of a product you can get from given reactants.

Excess Reactant

The excess reactant is the substance that remains after a chemical reaction has completed. It's the opposite of the limiting reactant.

In our example, \(\text{H}_2\text{O}\) is in excess because there is more available than what is needed to react with \(\text{CO}_2\). Initially, 3.55 moles of \(\text{H}_2\text{O}\) were available, but only 2.00 moles are required to completely react with 2.00 moles of \(\text{CO}_2\). This leaves us with a surplus.

Excess reactants can often be recycled or used in subsequent reactions. Calculating and predicting them help in optimizing material use and reducing waste. Understanding which reactant is in excess is just as important as knowing the limiting reactant, as it provides a fuller picture of the reaction's efficiency.

Molar Mass Calculation

Molar mass calculation is the process of finding the mass of one mole of any given substance, usually expressed in g/mol.

To find the molar mass, simply sum the atomic masses of all atoms in a chemical formula. For instance:

• \(\text{CO}_2\): Each carbon atom is about 12.01 g/mol, and each oxygen atom is about 16.00 g/mol, making the molar mass \(44.01\) g/mol.
• \(\text{H}_2\text{O}\): Hydrogen is about 1.01 g/mol each, and oxygen is about 16.00 g/mol, so the molar mass is \(18.02\) g/mol.
• \(\text{C}_6\text{H}_{12}\text{O}_6\): With 6 carbon, 12 hydrogen, and 6 oxygen atoms, it totals \(180.18\) g/mol.

These calculations are fundamental. They allow conversion between grams and moles, essential for stoichiometry, which is the quantitative relationship between reactants and products in a chemical reaction. Knowing the molar masses helps in accurate preparation and prediction of products in any given reaction.