Answer
2.164 g of pentane produces that quantity of $CO_2$ molecules.
Work Step by Step
When $C_5H_{12}$ is burned in excess oxygen, it produces $CO_2$ and water:
$C_5H_{12} + O_2 -- \gt CO_2 + H_2O$
1. Balance this reaction:
- Balance the $C$:
$C_5H_{12} + O_2 -- \gt 5CO_2 + H_2O$
- Balance the $H$:
$C_5H_{12} + O_2 -- \gt 5CO_2 + 6H_2O$
- Finally, balance the $O$:
Products: 5*2 + 6*1 = 16
Therefore, we should put "8" as the coefficient of $O_2$:
$C_5H_{12} + 8O_2 -- \gt 5CO_2 + 6H_2O$
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2. Calculate the molar mass of pentane:
Molar Mass ($C_5H_{12}$):
12.01* 5 + 1.008* 12 = 72.15g/mol
3. Use conversion factors to calculate the mass of pentane:
** $1 mol = 6.022 \times 10^{23} $
$9.033 \times 10^{22} (CO_2) \times \frac{1mol}{6.022 \times 10^{23}} \times \frac{1mol(C_5H_{12})}{5mol(CO_2)} \times \frac{72.15g(C_5H_{12})}{1mol(C_5H_{12})} = 2.164$ g $(C_5H_{12})$.
