Answer
The balanced chemical equation is shown below and 31.85 moles of $O_{2}$ are required to react with 4.9 moles of $C_{4}H_{10}$
Work Step by Step
$1C_{4}H_{10} + \frac{13}{2}O_{2} --> 4CO_{2} + 5H_{2}O$
4.9 moles $C_{4}H_{10}\times\frac{13/2 moles O_{2}}{1 mole C_{4}H_{10}} = 31.85 moles O_{2}$
