When 189.4 g of ethylene (C2H4) (molar mass= 28.05 g/mol) burns in oxygen to give carbon dioxide and water, how many grams of CO2 (molar mass= 44.01 g/mol) are formed?
C2H4 + 3O2 –> 2 CO2 + 2H2O
The Correct Answer and Explanation is :
When 189.4 grams of ethylene (C₂H₄) combust in oxygen, approximately 297.17 grams of carbon dioxide (CO₂) are produced.
Balanced Chemical Equation:
[ \text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} ]
Step-by-Step Calculation:
- Determine Moles of Ethylene (C₂H₄): [ \text{Moles of C}_2\text{H}_4 = \frac{\text{Mass of C}_2\text{H}_4}{\text{Molar Mass of C}_2\text{H}_4} = \frac{189.4\,\text{g}}{28.05\,\text{g/mol}} \approx 6.75\,\text{mol} ]
- Calculate Moles of CO₂ Produced: From the balanced equation, 1 mole of C₂H₄ produces 2 moles of CO₂. Therefore: [ \text{Moles of CO}_2 = 6.75\,\text{mol} \times 2 = 13.5\,\text{mol} ]
- Convert Moles of CO₂ to Grams: [ \text{Mass of CO}_2 = \text{Moles of CO}_2 \times \text{Molar Mass of CO}_2 = 13.5\,\text{mol} \times 44.01\,\text{g/mol} \approx 594.14\,\text{g} ]
Conclusion:
Combusting 189.4 grams of ethylene yields approximately 594.14 grams of carbon dioxide.
Explanation:
Combustion of ethylene involves a chemical reaction where ethylene (C₂H₄) reacts with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O). The balanced chemical equation for this reaction is:
[ \text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} ]
This equation indicates that one mole of ethylene reacts with three moles of oxygen to produce two moles of carbon dioxide and two moles of water.
Stoichiometric Calculations:
- Moles of Ethylene: The molar mass of ethylene (C₂H₄) is 28.05 g/mol. Therefore, the number of moles in 189.4 grams of ethylene is: [ \text{Moles of C}_2\text{H}_4 = \frac{189.4\,\text{g}}{28.05\,\text{g/mol}} \approx 6.75\,\text{mol} ]
- Moles of CO₂ Produced: According to the balanced equation, 1 mole of C₂H₄ produces 2 moles of CO₂. Therefore, 6.75 moles of C₂H₄ will produce: [ \text{Moles of CO}_2 = 6.75\,\text{mol} \times 2 = 13.5\,\text{mol} ]
- Mass of CO₂ Produced: The molar mass of CO₂ is 44.01 g/mol. Therefore, the mass of CO₂ produced is: [ \text{Mass of CO}_2 = 13.5\,\text{mol} \times 44.01\,\text{g/mol} \approx 594.14\,\text{g} ]
Conclusion:
Combusting 189.4 grams of ethylene produces approximately 594.14 grams of carbon dioxide.