Answer:
596 g of CO₂ is the mass formed
Explanation:
Combustion reaction:
C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g)
We determine moles of ethylene that has reacted:
189.6 g . 1mol / 28g = 6.77 moles
We assume the oxygen is in excess so the limiting reagent will be the ethylene.
1 mol of ethylene produce 2 moles of CO₂ then,
6.77 moles will produce the double of CO₂, 13.5 moles.
We convert the moles to mass: 13.5 mol . 44 g /1mol = 596 g
