Liquid octane (CH3(CH2)6CH3) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H20).

Question

Suppose 9.1 g of octane is mixed with 41.0 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Chien,

To determine how much Carbon dioxide could be produced by this chemical reaction, the first thing we need to do is write the balanced chemical equation. This will tell us how many moles of Octane and Oxygen gas react to form moles of Carbon dioxide and water. We can then identify which reactant will limit us from making the most Carbon dioxide possible based on the masses provided in this experiment scenario. Please note that I will be writing Octane as C₈H₁₈ rather than the structural formula you provided - this will make it easier for balancing.

Combustion Reaction: C₈H₁₈ + O₂ ---> CO₂ + H₂O

There are 8 Carbons on the left side of the reaction and only 1 on the right side. We can put a coefficient of 8 in front of the CO₂ to balance the C's. There are 18 H on the left and only 2 on the right, we can put a coefficient of 9 in front of the H₂O to fix this imbalance.

C₈H₁₈ + O₂ ---> 8 CO₂ + 9 H₂O

Both the Carbons and the Hydrogens are balanced, now let's do the Oxygen. There are 2 on the left side and 25 on the right side. We can fix this by placing a coefficient of 12.5 or 25/2 in front of the O₂.

C₈H₁₈ + 25/2 O₂ ---> 8 CO₂ + 9 H₂O

At this point, mathematically, the equation is balanced. In Chemistry, we want our coefficients to be whole numbers, so to fix this we can multiply the entire equation by 2 to turn the 25/2 into 25. Notice that we now have double the coefficients of everything.

2 C₈H₁₈ + 25 O₂ ---> 16 CO₂ + 18 H₂O

This balanced reaction tells us that 2 moles of octane will react with 25 moles of oxygen gas to produce 16 moles of carbon dioxide and 18 moles of water. Now we will determine how many moles of reactants we have based on the masses provided. To convert grams into moles, we will use the molecular weight of each compound.

9.1 g C₈H₁₈ x 1 mole C₈H₁₈ = 0.080 mol C₈H₁₈ available

114.23 g C₈H₁₈

41.0 g O₂ x 1 mole O₂ = 1.28 mol O₂ available

32.0 g O₂

The coefficients in our balanced equation tell us that we need to maintain a ratio of 25:2 O₂:C₈H₁₈

0.08 mol C₈H₁₈ x 25 mol O₂ = 1 mol O₂ needed (we have enough)

2 mol C₈H₁₈

1.28 mol O₂ x 2 mol C₈H₁₈ = 0.1024 mol C₈H₁₈ needed (more than we have) "limiting reactant"

25 mol O₂

C₈H₁₈ is the limiting reactant in this scenario because there is less moles of it available to fully react than the O₂. We will use all of the 0.080 mol of C₈H₁₈ and react it with only 1 mole of O₂. We can use the equation's coefficients as a ratio to determine how many moles of CO₂ will be produced by 0.080 mol C₈H₁₈

0.080 mol C₈H₁₈ x 16 mol CO₂ = 0.64 mol CO₂

2 mol C₈H₁₈

Answer 2

The first step to answering this problem is to write out the balanced equation

2 C₈H₁₈ + 17 O₂ --> 8 CO₂ + 18 H₂O

the rest in now a limiting reagent problem, you have to converrt the mass of octane and oxygen to their respectives moles and then to moles of carbon dioxide and then to grams of carbon dioxide

for octane:

9.1g X (1mol octane/114.23 g) X (8mol CO2/ 2 mole C8H18) X (44g CO2/1 mol CO2) = 14.02 g

for O2

41g O2 X ( 1 mol O2/ 32 g) X (8 mol CO2/17mol O2) X (44gCO2/1 mol)= 26.53 g

octane is therefore our limiting reagent so the final answer is 14 g (2 sig figs)