## Chemistry 10th Edition

The mass of propane that will produce 7.25 moles of water is equal to: $79.9g$
- Balance the reaction: $C_3H_8 + O_2 -- \gt CO_2 + H_2O$ - Balance the $C$: $C_3H_8 + O_2 -- \gt 3CO_2 + H_2O$ - Balance the $H$: $C_3H_8 + O_2 -- \gt 3CO_2 + 4H_2O$ - Balance the $O$: Products: 3*2 + 4*1 = 10 oxygens: $C_3H_8 + 5O_2 -- \gt 3CO_2 + 4H_2O$ -------------- 1. Calculate the molar mass of propane: $C_3H_8$: 12.01* 3 + 1.008* 8 = 44.09g/mol 2. Use the coefficients and the molar mass as conversion factors. $7.25 mol(H_2O) \times \frac{1mol(C_3H_8)}{4mol(H_2O)} \times \frac{44.09g(C_3H_8)}{1mol(C_3H_8)} = 79.9g(C_3H_8)$