Question

The combustion reaction for pentane is: c5h12 (l) + 8 O₂ (g) –> 5 CO₂⁺ (g) + 6 H₂O (g). If 933 g of CO₂⁺ is produced from the combustion of 6.45 moles of pentane with excess O₂, what is the mass of H₂O produced?

Answer 1

The mass of H2O produced from the combustion of 6.45 moles of pentane can be calculated using the stoichiometry of the reaction. Based on the molar ratio between CO2+ and H2O, the mass of H2O is determined by multiplying the moles of CO2+ produced by 6/5 and then converting the moles of H2O to grams using the molar mass of H2O. The mass of H2O produced is 4556.28 g.

Step-by-step explanation:

The combustion reaction for pentane is: C5H12 (l) + 8 O2 (g) → 5 CO2 (g) + 6 H2O (g).

In the reaction, 6.45 moles of pentane produce 933 g of CO2+. To find the mass of H2O produced, you can use the stoichiometry of the reaction. From the balanced equation, you can see that for every 6 moles of pentane burned, 6 moles of H2O are produced. Therefore, the molar ratio between CO2+ and H2O is 5:6.

First, calculate the number of moles of H2O produced using the molar ratio:

Moles of H2O = (moles of CO2+)*(6/5)

Then, convert the moles of H2O to grams using the molar mass of water (H2O):

Mass of H2O = (moles of H2O)*(molar mass of H2O)

Substituting the given values:

Mo1es of H2O = (933 g of CO2+)/(44.01 g/mol) * (6/5) = 252.91 moles

Mass of H2O = (252.91 moles)*(18.02 g/mol) = 4556.28 g