Problem 38 - Chemistry Matter and Change

Question

How many liters of propane gas \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) will undergo complete combustion with 34.0 \(\mathrm{L}\) of oxygen gas?

Step-by-Step Solution

Answer

6.8 L of propane is needed for the complete combustion with 34.0 L of oxygen gas.

Step 1: Write the Balanced Chemical Equation

First, identify the chemical equation for the combustion of propane. The balanced equation is: \[ \mathrm{C}_3\mathrm{H}_8 + 5\mathrm{O}_2 \rightarrow 3\mathrm{CO}_2 + 4\mathrm{H}_2\mathrm{O} \]. This shows that 1 mole of propane combusts with 5 moles of oxygen gas.

Step 2: Relate Volumes to Moles Using the Balanced Equation

Since gases at the same temperature and pressure have volumes proportional to moles, the ratio of volumes will match the ratio of coefficients in the balanced equation. For every 1 volume of \(\mathrm{C}_3\mathrm{H}_8\), 5 volumes of \(\mathrm{O}_2\) are required.

Step 3: Calculate Required Volume of Propane

Using the ratio from the balanced equation, determine how much propane is needed for 34.0 L of oxygen: \[ \text{Volume of } \mathrm{C}_3\mathrm{H}_8 = \frac{34.0 \text{ L of } \mathrm{O}_2}{5} = 6.8 \text{ L of } \mathrm{C}_3\mathrm{H}_8. \] Therefore, 6.8 L of propane are needed.

Key Concepts

Balanced Chemical Equation

In chemical reactions, equations must be balanced to indicate the conservation of mass. This means having the same number of each type of atom on both sides of the equation. This reflects the principle that matter can neither be created nor destroyed in a chemical reaction. Consider the balanced equation for the combustion of propane:

• Propane (\(\mathrm{C}_3\mathrm{H}_8\)) reacts with oxygen (\(\mathrm{O}_2\)) to produce carbon dioxide (\(\mathrm{CO}_2\)) and water (\(\mathrm{H}_2\mathrm{O}\)).
• The balanced equation is \(\mathrm{C}_3\mathrm{H}_8 + 5\mathrm{O}_2 \rightarrow 3\mathrm{CO}_2 + 4\mathrm{H}_2\mathrm{O}\).

The coefficients in the equation show that 1 mole of propane requires 5 moles of oxygen. The balanced equation is crucial for correctly predicting the amounts of reactants and products involved in the reaction.

Combustion Reactions

Combustion is a type of chemical reaction in which a substance reacts rapidly with oxygen to produce heat and light. In the case of propane, a hydrocarbon, combustion involves breaking the chemical bonds in the propane and oxygen molecules.

• Combustion reactions are exothermic, meaning they release energy.
• The general form for a hydrocarbon combustion reaction is \(\text{hydrocarbon} + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O}\).
• Complete combustion of propane (\(\mathrm{C}_3\mathrm{H}_8\)) ensures all carbon atoms form carbon dioxide, and hydrogen atoms form water.

Understanding combustion reactions help predict the energy produced and the products formed from burning fuels like propane.

Gas Laws

Gas laws describe how gases behave and relate to their volume, pressure, and temperature. Most importantly for stoichiometric calculations involving gas reactions is understanding the relationship between volume and moles at constant conditions.

• Boyle's Law, Charles's Law, and Gay-Lussac's Law are some basic gas laws.
• For stoichiometry in gases, Avogadro's Law is particularly important: it states that equal volumes of gases at the same temperature and pressure contain an equal number of moles.
• This principle allows us to use volume ratios directly as mole ratios if the conditions are constant.

Using gas laws, we simplify calculations for stoichiometry problems by relating moles directly to volumes.

Volume-Mole Relationship in Gases

The volume-mole relationship is a direct result of Avogadro's Law, which applies to reactions involving gases. At constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of gas.

• In the balanced equation for propane combustion, the ratio of volumes of gases is the same as the ratio of moles.
• This means 1 liter of \(\mathrm{C}_3\mathrm{H}_8\) requires 5 liters of \(\mathrm{O}_2\), matching the 1:5 mole ratio.
• To find the volume of another gas participating in the reaction, you just need to follow the mole ratio using the given volume.

This relationship simplifies calculations, allowing predictions of gas volumes needed or produced without having to measure the moles first. Understanding this concept is crucial for solving problems like determining the volume of propane needed for a given volume of oxygen.