How many grams of water will I obtain from the combustion of 0.9249 mol of pentane (C5H12) in the following reaction? C5H12 + O2 --> CO2 + H2O (unbalanced)

C5H12 + 8 (O2)--> 5 (CO2) + 6 (H2O)

*Molar Ratio of Pentane and water is 1:6

*Number of mol of H2O = 0.9249 ×6 = 5.5494 mol

* Molar mass of H2O = 18 g/mol

* Mass of H2O needded = 5.5494 mol × 18 g/mol

= 99.8892 g

Explanation:

* By balancing the equation we can get the stiochemetry of the reaction (Molar Ratio).

* Then we can find the number of mol of H2O neede relative to the no. of mol of Pentane.

* Multiply the mol of H2O by the molar mass of H2O to get the mass.