Give the balanced chemical equation for Cellular Respiration (all balanced chemical equations require states for full credit)
Solid glucose (C6H12O6) & oxygen gas react to product carbon dioxide and water vapor
Find the amount of carbon dioxide produced (in g) from the conversion of 60. g glucose assuming excess oxygen gas.
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Calculate the amount of carbon dioxide produced (in g) given the same amount of 60. g glucose, but 60. g oxygen gas as well.Explain why your answer changed or didn’t change.
The Correct Answer and Explanation is :
Let’s go through this step by step.
Step 1: Write the Balanced Chemical Equation
The reaction for cellular respiration is:
[
\text{C}6\text{H}{12}\text{O}_6 (s) + 6\text{O}_2 (g) \rightarrow 6\text{CO}_2 (g) + 6\text{H}_2\text{O} (g)
]
Each element is balanced:
- Carbon: 6 on both sides
- Hydrogen: 12 on both sides
- Oxygen: 18 on both sides (6 from glucose and 12 from oxygen gas)
Step 2: Find the Amount of CO₂ Produced from 60.0 g of Glucose (Excess O₂)
Convert Glucose to Moles
Molar mass of glucose ((\text{C}6\text{H}{12}\text{O}_6)):
[
(6 \times 12.01) + (12 \times 1.008) + (6 \times 16.00) = 180.16 \, g/mol
]
[
\frac{60.0 \, g}{180.16 \, g/mol} = 0.333 \, \text{mol C}6\text{H}{12}\text{O}_6
]
Use Stoichiometry to Find Moles of CO₂
From the balanced equation, 1 mole of glucose produces 6 moles of CO₂:
[
0.333 \times 6 = 2.00 \, \text{mol CO}_2
]
Convert Moles of CO₂ to Grams
Molar mass of CO₂:
[
(1 \times 12.01) + (2 \times 16.00) = 44.01 \, g/mol
]
[
2.00 \times 44.01 = 88.0 \, g \text{ CO}_2
]
Final Answer (with 3 significant figures):
[
\mathbf{88.0 \, g \text{ CO}_2}
]
Step 3: Find the Amount of CO₂ Produced with 60.0 g Glucose and 60.0 g O₂
Convert Oxygen to Moles
Molar mass of O₂:
[
2 \times 16.00 = 32.00 \, g/mol
]
[
\frac{60.0 \, g}{32.00 \, g/mol} = 1.875 \, \text{mol O}_2
]
Determine the Limiting Reactant
From the balanced equation, 1 mole of glucose requires 6 moles of O₂.
[
0.333 \times 6 = 2.00 \, \text{mol O}_2 \text{ required}
]
Since we only have 1.875 moles of O₂, it is the limiting reactant.
Find Moles of CO₂ from O₂
From the reaction, 6 moles of O₂ produce 6 moles of CO₂:
[
1.875 \times \frac{6}{6} = 1.875 \, \text{mol CO}_2
]
Convert to Grams of CO₂
[
1.875 \times 44.01 = 82.5 \, g \text{ CO}_2
]
Final Answer (with 3 significant figures):
[
\mathbf{82.5 \, g \text{ CO}_2}
]
Step 4: Explanation (Why Did the Answer Change?)
In the first case, glucose was the limiting reactant, and oxygen was in excess. This meant that all 60.0 g of glucose was fully consumed, producing 88.0 g of CO₂.
However, in the second case, oxygen became the limiting reactant because there wasn’t enough of it to react with all the glucose. Instead of 2.00 moles of CO₂, only 1.875 moles of CO₂ could form, resulting in a lower 82.5 g of CO₂.
This demonstrates the concept of limiting reactants in stoichiometry—when one reactant runs out first, the reaction stops, and the maximum amount of product depends on the limiting reactant. In real-world applications, ensuring sufficient reactants is essential for maximizing product yield.