Question

What is the limiting reagent when 56 grams of C2H4 react with 96 grams of O2 to produce 88 grams of CO2.

C2H4 + 3O2
2CO2 + 2H2O

Group of answer choices

C2H4

There is not a limiting reagent because the reaction involves stoichiometric amounts of each substance.

O2

CO2

Answer 1

To find the limiting reagent, we must first determine the moles of each reactant (C2H4 and O2) and then compare how much of each reactant is required based on the stoichiometry of the reaction.

**Step 1: Calculate moles of each reactant**

1. **C2H4 (Ethylene)**
- Molar mass of C2H4 = (2 × 12.01 g/mol) + (4 × 1.008 g/mol) = 28.05 g/mol
- Moles of C2H4 = mass / molar mass = 56 g / 28.05 g/mol = 1.996 moles (approximately 2 moles)

2. **O2 (Oxygen)**
- Molar mass of O2 = 2 × 16.00 g/mol = 32.00 g/mol
- Moles of O2 = mass / molar mass = 96 g / 32.00 g/mol = 3 moles

**Step 2: Use the balanced equation to find the required amounts of each reactant**

The balanced chemical equation is:
\[ \text{C2H4} + 3\text{O2} \rightarrow 2\text{CO2} + 2\text{H2O} \]

According to the balanced equation:
- 1 mole of C2H4 reacts with 3 moles of O2.

**Step 3: Determine how much O2 is needed for the available C2H4**
- With 2 moles of C2H4, the amount of O2 needed = 2 moles C2H4 × 3 moles O2/mole C2H4 = 6 moles O2.

**Step 4: Compare the available amount of O2 to what is needed**
- Available O2 = 3 moles
- Required O2 = 6 moles

Since we only have 3 moles of O2, which is less than the 6 moles required, O2 is the limiting reagent.

**Conclusion**
The limiting reagent in the reaction is **O2**.