Answer:
1.17 grams
Explanation:
Let's consider the balanced equation for the combustion of ethylene.
C₂H₄(g) + 3 O₂(g) → 2 CO₂(g) + 2 H₂O(l)
We can establish the following relations:
The grams of C₂H₄ burned to give 59.0 kJ of heat (q = -59.0 kJ) is:
[tex]-59.0kJ.\frac{1molC_{2}H_{4}}{-1411kJ} .\frac{28.05gC_{2}H_{4}}{1molC_{2}H_{4}} =1.17gC_{2}H_{4}[/tex]