Answer:
[tex]\large\boxed{\text{28 mol of O$_{2}$}}[/tex]
Explanation:
2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O
n/mol: 4.3
13 mol of O₂ react with 2 mol of 2C₄H₁₀
[tex]\text{Moles of O}_{2} = \text{4.3 mol C$_{4}$H$_{10}$} \times \dfrac{\text{13 mol O}_{2}}{\text{2 mol C$_{4}$H$_{10}$}} = \textbf{28 mol O}_{2}\\\\\text{The reaction requires $\large\boxed{\textbf{28 mol of O$_{2}$}}$}[/tex]
