If 6.02 grams of pentane is combusted with 7.52 grams of oxygen, how many grams of water are produced ?

Answer 1

First, write a correctly balanced equation for the reaction:

C₅H₁₂ + 8O₂ ==> 5CO₂ + 6H₂O

Next, find the limiting reactant:

6.02 g C₅H₁₂ x 1 mole/72.15 g = 0.0834 moles C₅H₁₂

7.52 g O₂ x 1 mole/32 g = 0.235 moles O₂

Since it takes 8 moles O₂ for each mole of C₅H₁₂, O₂ is limiting

Finally, find the grams of H₂O that can be produced from 0.235 moles O₂:

0.235 moles O₂ x 6 moles H₂O/8 moles O₂ x 18 g H₂O/mole = 3.17 g H₂O produced