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)

Mass = 2.7 g

Explanation:

Given data:

Number of moles of C₅H₁₂ = 0.9249 mol

Mass of water = ?

Solution:

Chemical equation:

C₅H₁₂ + 8O₂     →     5CO₂ + 6H₂O

Now we will compare the moles of water with pentane.

                 C₅H₁₂       :         H₂O

                    1            :           6

                0.9249     :          1/6×0.9249= 0.15 mol

Mass of water:

Mass = number of moles × molar mass

Mass = 0.15 mol × 18 g/mol

Mass = 2.7 g