Limiting Reactants

Question

Liquid hexane CH3CH24CH3

 will react with gaseous oxygen O2

 to produce gaseous carbon dioxide CO2

 and gaseous water H2O

. Suppose 6.9 g of hexane is mixed with 31.9 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 2

 significant digits

Answer 1

You provided the formula of the compound as CH3CH24CH3 and you should have written CH3(CH2)₄CH3. This would then, in fact, be hexane.

Now, write the equation for the combustion of hexane...

CH3(CH2)₄CH3 + O2 ==> CO2 + H2O ... unbalanced

In order to balance this, It's easier to write it as follows:

2C₆H₁₄ + 19O₂ ==> 12CO₂ + 14H₂O ... balanced equation

An easy way to find the limiting reactant is to divided the moles of each reactant by the coefficient in the balanced equation. Thus..

For C₆H₁₄: 6.9 g x 1 mol/86.18 g = 0.0801 moles C₆H₁₄ (÷2->0.040)

For O2: 31.9 g x 1 mol/32 g = 0.997 moles O2 (÷19->0.05)

Since 0.04 is < 0.05, C₆H₁₄ is LIMITING and will determine the amount of product that can form

0.0801 mols C₆H₁₄ x 14 mols H₂O / 2 mols C₆H₁₄ x 18 g H₂O/mol H₂O = 10.09 g H2O formed = 10. g H2O (2 sig. figs.)

Answer 2

You need the reaction equation in order to determine the LR. The hydrocarbon that you gave is propane, not hexane. (Hexane is C₆H₁₄

You need to write a balanced equation so you can use the stoichiometric coefficients.

Since you want to calculate the mass of product, water, anyway, the simplest way to find the answer is just to grind out the water formed from each reactant as if the other one was in excess. Whichever one give less product, that's the limiting reactant and you have your answer for the water produced.

g of O₂ * (mole/g of O₂) *( moles water/mole O₂)*(g /mole of water) = g of water

You get g/mole is molar mass and the mole ratio is from the stochiometric coefficients in your reaction.

Do the same thing for the fuel