Liquefied petroleum gas (LPG) consists primarily of propane, $\mathrm{C}_{3} \mathrm{H}_{8}(l)\( or butane \)\mathrm{C}_{4} \mathrm{H}_{10}(l)$ (a) Write a balanced chemical equation for the complete combustion of propane to produce \(\mathrm{CO}_{2}(g)\) as the only carbon-containing product. (b) Write a balanced chemical equation for the incomplete combustion of propane to produce \(\mathrm{CO}(g)\) as the only carbon-containing product. (c) At \(25^{\circ} \mathrm{C}\) and \(101.3 \mathrm{kPa}\) pressure, what is the minimum quantity of dry air needed to combust \(10.0 \mathrm{~mL}\) of \(\mathrm{C}_{3} \mathrm{H}_{8}(l)\) completely to \(\mathrm{CO}_{2}(g)\) ? The density of the LPG is $0.50 \mathrm{~g} / \mathrm{mL}$.

Step by step solution

01

Balanced Equation for Complete Combustion of Propane

To balance the complete combustion of propane, we need to produce CO2 and H2O as products. The equation is: \[C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O\]

02

Balanced Equation for Incomplete Combustion of Propane

To balance the incomplete combustion of propane, we need to produce CO and H2O as products. The equation is: \[2C_3H_8 + 7O_2 \rightarrow 4CO_2 + 2CO + 8H_2O\]

03

Calculate the Mass of the Propane Sample

We are given that 10.0 mL of propane has a density of 0.50 g/mL. To find the mass of the propane sample, we multiply the volume by the density: Mass of propane = Volume × Density Mass of propane = 10.0 mL × 0.50 g/mL = 5.0 g

04

Moles of Propane

To find the moles of propane, we divide the mass by the molar mass (the molecular weight of propane, C3H8, is 44.10 g/mol): Moles of propane = Mass of propane / Molar mass of propane Moles of propane = 5.0 g / 44.10 g/mol = 0.1134 mol

05

Moles of O2 Required for Complete Combustion

Using stoichiometry, we can find the moles of O2 required for complete combustion. From the balanced equation for complete combustion, 1 mole of propane requires 5 moles of O2: Moles of O2 required = Moles of propane × (5 moles O2 / 1 mole propane) Moles of O2 required = 0.1134 mol × 5 = 0.5670 mol

06

Find the Volume of Air Needed

Dry air is about 21% oxygen by volume. We can use the ideal gas law (PV = nRT) to find the volume of oxygen required at the given conditions (T = 25°C = 298 K, P = 101.3 kPa), and then divide by the percentage of oxygen in the air to find the volume of air needed. R (ideal gas constant) = 8.314 kJ/(k . mol) Volume of O2 = n O2 × R × T / P Volume of O2 = 0.5670 mol × 8.314 kJ/(k . mol) × 298 K / 101.3 kPa = 0.01242 m³ Now, we need to find the volume of air needed: Volume of air = Volume of O2 / 0.21 (since air is 21% O2) Volume of air = 0.01242 m³ / 0.21 = 0.05914 m³ So, the minimum quantity of dry air needed to combust 10.0 mL of propane completely to CO2 gas is 0.05914 m³.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!