Final answer:
When 189.6 g of ethylene burns in oxygen, 594.8 grams of CO₂ are formed. This is calculated by first converting the mass of ethylene to moles, then using the stoichiometry of the balanced chemical equation to find the moles of CO₂ produced.
Step-by-step explanation:
The combustion reaction of ethylene (C₂H₄) is as follows:
C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g)
First, we need to convert the mass of ethylene to moles using its molar mass. The molar mass of ethylene is approximately 28.05 g/mol:
189.6 g C₂H₄ × (1 mol C₂H₄ / 28.05 g) = 6.76 mol of C₂H₄
According to the balanced equation, 1 mole of C₂H₄ produces 2 moles of CO₂. So, 6.76 moles of C₂H₄ will produce:
6.76 mol C₂H₄ × (2 mol CO₂ / 1 mol C₂H₄) = 13.52 mol CO₂
Now, we can calculate the mass of CO₂ produced using its molar mass (44.01 g/mol):
13.52 mol CO₂ × (44.01 g/mol) = 594.8 g CO₂
Therefore, the combustion of 189.6 g of ethylene will produce 594.8 grams of CO₂.