Question

Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 63. g of octane is mixed with 59.4 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to significant digits.

Answer 1

The mass of carbon dioxide produced is 52.3 grams

Step-by-step explanation:

Step 1: Data given

Mass of octane = 63.0 grams

Mass of oxygen = 59.4 grams

Molar mass of octane = 114.23 g/mol

Molar mass of oxygen = 32.0 g/mol

Step 2: The balanced equation

2C8H18 + 25O2 → 16CO2 + 18H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles octane = 63.0 grams / 114.23 g/mol

Moles octane = 0.5515 moles

Moles oxygen = 59.4 grams / 32.0 g/mol

Moles oxygen = 1.856 moles

Step 4: Calculate limiting reactant

For 2 moles octane we need 25 moles oxygen to produce 16 moles CO2 and 18 moles H2O

Oxygen is the limiting reactant. IT will completely be consumed (1.856 moles). Octane is in excess. There will react 1.856 / 12.5 = 0.1485 moles

There will remain 0.5515 -0.1485 = 0.403 moles octane

Step 5: Calculate mass of CO2

For 2 moles octane we need 25 moles oxygen to produce 16 moles CO2 and 18 moles H2O

For 1.856 moles O2 we'll have 16/25 * 1.856 = 1.1878 moles CO2

Step 6: Calculate mass of CO2

Mass of CO2 = moles CO2 * molar mass CO2

Mass of CO2 = 1.1878 moles * 44.01 g/mol

Mass of CO2 = 52.3 grams

The mass of carbon dioxide produced is 52.3 grams

Answer 2

52.1 g is the maximum mass of CO₂, that can be produced by this combustion

Step-by-step explanation:

Mass of Octane: 63 g

Mass of O₂: 59.4 g

This is a combustion reaction where the products, are always water and CO₂. We define the equation:

2C₈H₁₈ (l) + 25O₂(g) → 16CO₂(g) + 18H₂O(g)

As we have, both mases of each reactant, we must define which is the limiting reagent. We convert the mass to moles:

63 g. 1mol / 114g = 0.552 moles

59.4 g . 1mol / 32g = 1.85 moles

Certainly, the limiting reagent is the oxygen:

2 moles of octane need 25 moles of O₂ to react

Therefore, 0.552 moles of octane must need (0.552 . 25) /2 = 6.9 moles of O₂ (I do not have enough moles of oxygen, I need 6.9 and I only got 1.85 moles)

When we know the limiting reagent we can do the calculations with the stoichiometry of the reaction:

25 moles of O₂ can produce 16 moles of CO₂

Therefore, 1.85 moles of O₂ may produce (1.85 . 16) /25 = 1.18 moles.

We convert the moles to mass to get the final answer:

1.18 mol . 44 g / 1mol = 52.1 g