Gaseous C2H4 reacts with O2according to the following: C2H4(g) + 3O2(g) → 2CO2(g) + 2 H2O(g) What volume of oxygen at STP is needed to react with 1.50 mole of C2H4?
The correct answer and explanation is:
Correct Answer: 1.50 mol C₂H₄ × (3 mol O₂ / 1 mol C₂H₄) × (22.4 L / 1 mol O₂) = 100.8 L of O₂ at STP
Detailed Explanation (Approx. 300 words):
The chemical reaction given is: C2H4(g)+3O2(g)→2CO2(g)+2H2O(g)\text{C}_2\text{H}_4(g) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 2\text{H}_2\text{O}(g)
This is a balanced combustion reaction of ethene (C₂H₄). It shows that 1 mole of ethene reacts with 3 moles of oxygen gas to produce carbon dioxide and water vapor.
In this problem, you’re asked to find the volume of oxygen gas at STP (Standard Temperature and Pressure) needed to react with 1.50 moles of C₂H₄.
Step 1: Use the Mole Ratio
From the balanced equation: 1 mol C2H4:3 mol O21\ \text{mol C}_2\text{H}_4 : 3\ \text{mol O}_2
So if you have 1.50 mol of C₂H₄: 1.50 mol C2H4×3 mol O21 mol C2H4=4.50 mol O21.50\ \text{mol C}_2\text{H}_4 \times \frac{3\ \text{mol O}_2}{1\ \text{mol C}_2\text{H}_4} = 4.50\ \text{mol O}_2
Step 2: Convert Moles of O₂ to Volume at STP
At STP (0°C and 1 atm), 1 mole of any ideal gas occupies 22.4 liters. 4.50 mol O2×22.4 L1 mol=100.8 L4.50\ \text{mol O}_2 \times \frac{22.4\ \text{L}}{1\ \text{mol}} = 100.8\ \text{L}
Final Answer: 100.8 liters of O₂
This means that to completely combust 1.50 moles of ethene, you would need 100.8 liters of oxygen gas at standard temperature and pressure.
This calculation illustrates how stoichiometry and molar gas volume principles are applied to real-world chemical reactions. In chemistry, such conversions are crucial in predicting gas volumes in industrial or laboratory settings, ensuring the correct proportions of reactants are used for efficient and safe reactions.