Question

Combustion of hydrocarbons such as butane () 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. Write a balanced chemical equation, including physical state symbols, for the combustion of gaseous butane into gaseous carbon dioxide and gaseous water. 2. Suppose 0.360 kg of butane are burned in air at a pressure of exactly 1 atm and a temperature of 20.0 °C. Calculate the volume of carbon dioxide gas that is produced Be sure your answer has the correct number of significant digits.

Answer 1

Answer:

1. C₄H₁₀ + ¹³/₂O₂ → 4CO₂ + 5H₂O

2. V = 596L

Explanation:

Butane (C₄H₁₀) reacts with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O) thus:

C₄H₁₀ + O₂ → CO₂ + H₂O

1. The balanced chemical equation is:

C₄H₁₀ + ¹³/₂O₂ → 4CO₂ + 5H₂O

2. 0,360kg of butane are:

360g×$\frac{1mol}{58,12g}$=6,19moles of butane

These moles of butane are:

6,19moles of butane×$\frac{4CO_2}{1molButane}$= 24,8 moles CO₂

Using V=nRT/P

Where:

n are moles (24,8 moles CO₂); R is gas constant (0,082atmL/molK); T is temperature, 20°C (293,15K); and P is pressure (1atm).

Volume (V) is:

V = 596L

I hope it helps!