Given the following unbalanced equation, calculate the liters of water produced in this combustion reaction if you react 1 mole of CasH3o with enough oxygen. The density of water is 1.0g / mL.

Question

Given the following unbalanced equation, calculate the liters of water

produced in this combustion reaction if you react 1 mole of CasH3o with

enough oxygen. The density of water is 1.0g / mL.

C¹⁵H₃₀+ O₂H₂O + CO₂

Answer 1

To balance a combustion reaction, you usually balance the C, then the H, and then the O after assuming the H,C,O fuel has a 1 stoichiometric coefficient. Half of the time, you will have n/2 in front of the O₂ which I think is preferable as it corresponds to the appropriate equation per mole of fuel (Enthalpy will be the molar heat of combustion), but most books and standardized tests require integer stoichiometry - just multiply all terms by 2.

C₁₅H₃₀ + 45/2 O₂ → 15CO₂ + 15H₂O

Now, it's just a stoichiometry problem going from extensive to extensive variable. Cancel units to get where you want to go and put in the appropriate conversions and information to get there:

1 mole fuel (15 mole W / 1 mole fuel) (18 g W/mole W) (1 ml /1 g W) (1 liter/1000 ml) = liters of water.

Answer 2

Looks like you mixed 2 problems together. You have one with CasH3o which I have no idea what that's supposed to be, and then you have C15H30 + O2 = H2O + CO2 (I think). This latter one is the one I'll use.

2C₁₅H₃₀ + 45O₂ ==> 30H₂O + 30CO₂ ... balanced equation

1 mol C₁₅H₃₀ x 30 mol H₂O / 2 mol C₁₅H₃₀ = 15 moles H₂O produced

15 mols H₂O x 18 g H₂O / mol H₂O = 270 g H₂O produced

270 g H₂O x 1 ml / 1.0 g = 270 mls H₂O produced

270 mls H₂O x 1 L / 1000 mls = 0.27 liters of H₂O produced