Answer
1.64 mol $CO_{2}$.
Work Step by Step
Balanced chemical reaction:
2 $C_{4}H_{10}$(g) + 13 $O_{2}$ (g) ${\rightarrow}$ 8 $CO_{2}$ (g) + 10 $H_{2}O$ (g)
From this balanced chemical equation, we see that 2 moles $C_{4}H_{10}$ give 8 moles $CO_{2}$. So the conversion factor should be $\frac{8_{mole (CO_{2})}}{2_{mol (C_{4}H_{10})}}$.
Therefore the amount of moles $CO_{2}$ produced by 0.41 mol $C_{4}H_{10}$ , can be found:
0.41 mol $C_{4}H_{10}$ x $\frac{8_{mole (CO_{2}})}{2_{mol (C_{4}H_{10})}}$ = 1.64 mol $CO_{2}$
