Question

Combustion of hydrocarbons such as propane ( C3H8 ) produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's atmosphere can trap the Sun's heat, raising the average temperature of the Earth. For this reason there has been a great deal of international discussion about whether to regulate the production of carbon dioxide. 1. Write a balanced chemical equation, including physical state symbols, for the combustion of gaseous propane into gaseous carbon dioxide and gaseous water. 2. Suppose 0.150kg of propane are burned in air at a pressure of exactly 1atm and a temperature of 12.0°C . Calculate the volume of carbon dioxide gas that is produced. Round your answer to 3 significant digits.

Answer 1

Answer:

1. C₃H₈(g) + 5O₂ (g)  → 3CO₂ (g) + 4H₂O (g)

2. The volume of CO₂ produced is 238 l.

Explanation:

Hi there!

The chemical equation is the following:

C₃H₈(g) + 5O₂ (g)  → 3CO₂ (g) + 4H₂O (g)

The molar mass of propane is calculated as follows:

mass of 1 mol carbon = 12 g

mass of 1 mol hydrogen = 1 g

mass of 1 mol propane 3 · (12 g) + 8 · (1 g) = 44 g

If we have 150 g of propane, we will have (150 g · 1 mol / 44 g) 3.4 mol propane.

Looking at the chemical equation, notice that 1 mol propane produces 3 mol CO₂. Then 3.4 mol of propane will produce:

(3.4 mol C₃H₈ ·  3 mol CO₂ / mol C₃H₈ ) 10.2 mol CO₂

Using the Ideal Gas Law we can obtain the volume of CO₂ produced:

P · V = n · R · T

Where:

P = pressure.

V = volume

n = number of moles.

R = gas constant (0.082 atm · l / ( K · mol))

T = temperature in kelvin.

V = n · R · T / P

V = 10.2 mol · 0.082 atm · l/ (K mol) · 285 K / 1 atm

Notice that 12 °C = 273  + 12  = 285 K

V = 238 l

The volume of CO₂ produced is 238 l.

Have a nice day!