Question

The combustion of propane may be described by the chemical equation CH₃CH₂CH₃(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g). How many grams of O₂(g) are needed to completely burn 20.3 g C₂H₆(g)?

Option a: 77.4 g O²
Option b: 110.7 g O²
Option c: 154.8 g O²
Option d: 83.2 g O²

Answer 1

Using a stoichiometric calculation, 75.64 g of O2 is required to completely burn 20.3 g of C2H6, which closely matches Option a: 77.4 g O2.

Step-by-step explanation:

Calculating the Amount of Oxygen Needed for Complete Combustion of Propane

First, we will deal with the provided equation which is slightly incorrect as it refers to the combustion of ethane (C2H6), not propane (C3H8). However, since the question asks about the amount of O2 required to combust ethane, we will proceed with that calculation, and it's important to correct the molecular formula of ethane: C2H6(g).

The balanced equation for the combustion of ethane (C2H6) is:

2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)

To find out how much O2 is needed to burn 20.3 g of C2H6, we'll use stoichiometry:

1. Calculate the molar mass of C2H6: (2 × 12.01) + (6 × 1.008) = 30.07 g/mol

2. Convert grams of C2H6 to moles: 20.3 g C2H6 ÷ 30.07 g/mol = 0.675 moles C2H6

3. Determine moles of O2 needed using the balanced equation: (7 moles O2 / 2 moles C2H6) × 0.675 moles C2H6 = 2.36375 moles O2

4. Convert moles of O2 to grams: (2.36375 moles O2) × (32.00 g/mol O2) = 75.64 g O2

Based on calculations, the mass of O2 needed is 75.64 g, which doesn't exactly match any of the given options. However, Option a is closest to our calculated value.