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. L

Answer 1

Answer:

239 L

Explanation:

1. Propane reacts with oxygen to produce carbon dioxide and water:

$C_3H_8 (g) + O_2 (g)\rightarrow CO_2 (g) + H_2O (l)$

Firstly, 3 carbon atoms are required on the right:

$C_3H_8 (g) + O_2 (g)\rightarrow 3 CO_2 (g) + H_2O (l)$

Secondly, 8 hydrogens in total (4 water molecules) are required on the right:

$C_3H_8 (g) + O_2 (g)\rightarrow 3 CO_2 (g) + 4 H_2O (l)$

On the right, we have a total of 10 oxygen atoms, this implies we need 5 oxygen molecules on the left:

$C_3H_8 (g) + 5 O_2 (g)\rightarrow 3 CO_2 (g) + 4 H_2O (l)$

2. Calculate moles of propane using the the ratio of mass to molar mass:

$n_1 = \frac{m_1}{M_1} = \frac{150 g}{44.1 g/mol} = 3.40 mol$

According to the stoichiometry, we have 3 times greater amount of carbon dioxide:

$n_2 = 3n_2 = 3\cdot 3.40 mol = 10.2 mol$

Use the ideal gas law to solve for volume:

$pV = nRT\therefore V = \frac{nRT}{p} = \frac{10.2 mol\cdot 0.08206 \frac{L atm}{mol K}\cdot 285.15 K}{1 atm} = 239 L$