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 g of carbon dioxide and 64.0 g of water available for photosynthesis. a. Write the balanced chemical equation for the reaction. b. Determine the limiting reactant. c. Determine the excess reactant. d. Determine the mass in excess. e. Determine the mass of glucose produced.

Step-by-Step Solution

Answer

The balanced chemical equation for photosynthesis is \(6\mathrm{CO}_{2} + 6\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} + 6\mathrm{O}_{2}\). The limiting reactant is CO₂, and the excess reactant is H₂O. The mass in excess is 27.93 g of water. The mass of glucose produced is 60.06 g.

Step 1: Write the balanced chemical equation for photosynthesis

Photosynthesis is the process of plants and other autotrophs converting carbon dioxide (CO2) and water (H2O) into glucose (C6H12O6) and releasing oxygen (O2) as a byproduct. The balanced chemical equation for this process is: 6CO2 + 6H2O -> C6H12O6 + 6O2

Step 2: Determine the moles of reactants

We start by calculating the moles of each reactant: For Carbon dioxide (CO2): Molecular weight of CO2 = 12.01 (C) + 2 * 16.00 (O) = 44.01 g/mol Moles of CO2 = mass of CO2÷ molecular weight Moles of CO2 = 88.0 g ÷ 44.01 g/mol = 2.00 mol For Water (H2O): Molecular weight of H2O = 2 * 1.01 (H) + 16.00 (O) = 18.02 g/mol Moles of H2O = mass of H2O ÷ molecular weight Moles of H2O = 64.0 g ÷ 18.02 g/mol = 3.55 mol

Step 3: Determine the limiting and excess reactants

To determine the limiting reactant, we need to compare the ratio of moles of reactants to the stoichiometric ratio in the balanced equation: Stoichiometric ratio CO2 : H2O = 6 : 6 (from the balanced equation) We compare the ratios: CO2/H2O = (2.00 mol) / (3.55 mol) = 0.56 Because the ratio 0.56 is smaller than the stoichiometric ratio 1:1, CO2 is the limiting reactant, and H2O is the excess reactant. So, the answers are b. limiting reactant: CO2, and c. excess reactant: H2O.

Step 4: Determine the mass in excess

To find the mass in excess, we need to calculate the amount of water used in the reaction: From the stoichiometry, 6 moles of water react with 6 moles of CO2. We have 2.00 moles of CO2, so the amount of water used in the reaction is: 2.00 mol CO2 * (6 mol H2O / 6 mol CO2) = 2.00 mol H2O Now, we can calculate the remaining (unused) moles of water: 3.55 mol (initial) - 2.00 mol (used) = 1.55 mol H2O (remaining, unused) Finally, we can convert the remaining moles of water to mass: Mass of unused water = 1.55 mol * 18.02 g/mol = 27.93 g So, the mass in excess (d) is 27.93 g of water.

Step 5: Determine the mass of glucose produced

To calculate the mass of glucose (C6H12O6) produced, we start by finding the amount (moles) of glucose produced: Molecular weight of glucose = 6 * 12.01 (C) + 12 * 1.01 (H) + 6 * 16.00 (O) = 180.18 g/mol From the stoichiometry, 1 mole of glucose is produced for every 6 moles of CO2. We have 2.00 moles of CO2: (2.00 mol CO2) * (1 mol C6H12O6 / 6 mol CO2) = 1/3 mol C6H12O6 Now, we can calculate the mass of glucose produced: Mass of glucose = (1/3 mol C6H12O6) * 180.18 g/mol = 60.06 g So, the mass of glucose produced (e) is 60.06 g.

Key Concepts

Limiting Reactant

In chemical reactions, the limiting reactant is the substance that is used up first and thus prevents more product from being formed. It's akin to the bottleneck in a production line; once it runs out, the process halts. Determining the limiting reactant is vital in calculating the theoretical yield of a reaction.

To identify the limiting reactant, calculate the moles of each reactant and compare them to the required stoichiometric ratios. For photosynthesis, carbon dioxide (CO2) and water (H2O) are required in equal molar ratios. If the moles of CO2 available are less than the moles of H2O (relative to their stoichiometric ratio), CO2 becomes the limiting reactant, as seen in our example.

Excess Reactant

Conversely, the excess reactant is the one that remains after the reaction has ceased. There is more of the excess reactant than is necessary to completely react with the limiting reactant. Knowing the excess reactant is crucial for calculating any remaining materials after the reaction and for determining reaction efficiency.

In the context of photosynthesis, once we've established the limiting reactant, in this case, CO2, we can then determine that water (H2O) is the excess reactant, as there are leftover moles of H2O that won’t be used in the reaction.

Mole Concept

The mole concept is a bridge linking the mass of materials to the number of atoms, molecules, or formula units involved. One mole is Avogadro's number (\(6.022 \times 10^{23}\) entities) of anything, which is similar to a 'dozen' but for a very large quantity. It's the chemist’s shorthand for handling large quantities of small entities (like atoms or molecules).

Using the mole concept, we converted the mass of CO2 and H2O to moles in order to compare them directly and find the limiting and excess reactants. For instance, with 88.0 g of CO2, we used its molar mass to find that it equaled 2.00 moles of CO2.

Stoichiometry

The term stoichiometry comes from the Greek words 'stoicheion' (element) and 'metron' (measure). In essence, stoichiometry deals with the quantification of reactants and products in chemical reactions. It tells us the proportional relationships between reactants and products and serves as the foundation of calculations for chemical reactions.

In our photosynthesis example, stoichiometry allowed us to calculate the mass of glucose produced and the excess mass of water. Using the balanced chemical equation, we can see the ratios required for the reaction to occur and apply these to our reactants to find the outcome of the reaction.