Answer:
[tex]\boxed{\text{46.07 g}}[/tex]
Explanation:
M_r: 46.07
C₂H₅OH(ℓ) + 3O₂(g) ⟶ 2CO₂(g) + 3H₂O(ℓ); ΔH = -1418 kJ·mol⁻¹
[tex]\text{Mass} = \text{1418 kJ} \times \dfrac{\text{1 mol}}{\text{1418 kJ}} \times \dfrac{\text{46.07 g}}{\text{1mol}}= \textbf{46.07 g}\\\\\text{1418 kJ are equivalent to $\boxed{\textbf{46.07 g}}$ of ethanol}[/tex]