## Chemistry 10th Edition

2.164 g of pentane produces that quantity of $CO_2$ molecules.
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 $O_2$: $C_5H_{12} + 8O_2 -- \gt 5CO_2 + 6H_2O$ ----------------------- 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 $(C_5H_{12})$.