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The problem you're working on involves determining the maximum number of grams of product (CO₂) in a balanced chemical reaction by constructing a BCA (Before, Change, After) table based on stoichiometry. Let's break it down step by step.

### Given Balanced Reaction:
\[
\text{C}_2\text{H}_4\text{O}_2(g) + 2 \text{O}_2(g) \rightarrow 2 \text{CO}_2(g) + 2 \text{H}_2\text{O}(g)
\]

### Given Information:
- 45.2 mol of \(\text{C}_2\text{H}_4\text{O}_2\)
- 76.5 mol of \( \text{O}_2 \)

### Step-by-Step Stoichiometry:

1. **Limiting Reactant Determination**:
   First, we must determine the limiting reactant. The reaction requires 1 mol of \(\text{C}_2\text{H}_4\text{O}_2\) for every 2 mol of \( \text{O}_2 \). 

   Let's calculate the number of moles of \( \text{O}_2 \) required for 45.2 mol of \(\text{C}_2\text{H}_4\text{O}_2\):

   \[
   \text{Required moles of } O_2 = 45.2 \text{ mol } \times 2 = 90.4 \text{ mol of } O_2
   \]

   Since you only have 76.5 mol of \( \text{O}_2 \), oxygen is the limiting reactant.

2. **Construct the BCA Table**:

|                          | \(\text{C}_2\text{H}_4\text{O}_2(g)\) | \(O_2(g)\) | \(CO_2(g)\) | \(H_2O(g)\) |
|--------------------------|-----------------------------------|------------|-------------|-------------|
| **Before (mol)**          | 45.2                              | 76.5       | 0           | 0           |
| **Change (mol)**          | \(-x\)                            | \(-2x\)    | \(+2x\)     | \(+2x\)     |
| **After (mol)**           | \(45.2 - x\)                      | \(76.5 - 2x\) | \(2x\)    | \(2x\)      |

Since oxygen is the limiting reactant, let’s calculate the value of \(x\) using the amount of oxygen available.

\[
2x = 76.5 \implies x = 38.25
\]

3. **Determine Amount of Products**:

- Moles of \(CO_2\) produced: \(2x = 2 \times 38.25 = 76.5\) mol.
- Moles of \(H_2O\) produced: \(2x = 76.5\) mol.

4. **Mass of CO₂ Produced**:
To find the mass of \(CO_2\), use its molar mass:
\[
\text{Molar mass of CO}_2 = 12 + 16 \times 2 = 44 \, \text{g/mol}
\]
Mass of \(CO_2 = 76.5 \, \text{mol} \times 44 \, \text{g/mol} = 3366 \, \text{g}\).

### Final Answer:
The maximum mass of \(CO_2\) that can be produced is **3366 grams**.

Would you like further details, or do you have any questions about the process?

### Related Questions:
1. How do you determine the limiting reactant in any chemical reaction?
2. What happens when the amount of the limiting reactant is less than required?
3. Why is the molar mass of a compound important in stoichiometry?
4. How does the ratio of coefficients in a balanced reaction affect product formation?
5. What role does the BCA table play in solving stoichiometric problems?

### Tip:
Always check the units and the consistency of your mole ratios to avoid miscalculations in stoichiometric problems.

Determine the maximum number of grams of product that can be produced in the balanced reaction described below by constructing a BCA table and calculating the maximum moles of CO₂ product. A reaction with 45.2 mol of C₂H₄O₂ and 76.5 mol of O₂ occurred. Based on your knowledge of stoichiometry, set up the table to determine the amounts of each reactant and product after the reaction goes to completion.

Calculate the maximum grams of CO₂ produced using stoichiometry from a reaction involving 45.2 mol of C₂H₄O₂ and 76.5 mol of O₂. Learn how to construct a BCA table and identify the limiting reactant.

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