If 22.0 grams of pentane undergos combustion how many molecules of water are produced

Question

1C5H12 + 8O2 = 5CO2 + 6H2O

Answer 1

Hi Cam,

So there are multiple steps that you need to complete in order to get to your answer:

Step 1: Ensure the reaction equation is balanced (both the reactants and products side should have the same number of each atom)

Step 2: Determine your limiting reactant. Your limiting reactant is the reactant that is 100% consumed by the time the reaction is completed. You find the limiting reactant by converting the masses of your available reactants to moles, and then calculating the moles necessary to completely consume one reactant with the other reactant. Typically for these types of questions you should be given the masses of both reactants, not just one. So I'm going to assume that pentane is your limiting reactant since the mass of oxygen is not provided (if your problem does give you a mass for oxygen please comment on this answer and let me know so I can update my answer).

Step 3: Do your stoichiometric conversions to convert the mass of pentane to moles, and then relate the moles of pentane to moles of water produced (using the coefficients from your balanced equation), and then convert from moles to molecules using Avogadro's number.

22 grams pentane (1 mole pentane/72.15 grams)(6 moles water / 1 mole pentane)(6.02x10²³ molecules /1 mole water) = 1.1014 x 10²⁴ molecules of water are produced from this reaction