Question

Consider the following unbalanced chemical equation for the combustion of pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right) :\) $$ \mathrm{C}_{5} \mathrm{H}_{12}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ If 20.4 g of pentane are burned in excess oxygen, what mass of water can be produced, assuming 100\(\%\) yield?

Short answer

The mass of water produced when 20.4 g of pentane is burned in excess oxygen with a 100% yield is 26.15 g.

Step-by-step solution

Balance the chemical equation

To balance the chemical equation, we need to make sure that the same number of atoms of each element is present on both sides of the equation. The balanced chemical equation for the combustion of pentane is: $$ \mathrm{C}_{5} \mathrm{H}_{12}(l) + 8\mathrm{O}_{2}(g) \longrightarrow 5\mathrm{CO}_{2}(g) + 6\mathrm{H}_{2} \mathrm{O}(l) $$

Convert the mass of pentane to moles

Using the periodic table, we can calculate the molar mass of pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right)\) as follows: $$ M_{C_{5}H_{12}} = (5 \times 12.01 \,\text{g/mol}) + (12 \times 1.01 \,\text{g/mol}) = 72.05 + 12.12 = 84.17\, \text{g/mol} $$ Now, we can convert the mass of pentane (20.4 g) to moles: $$ \text{moles of pentane} = \frac{\text{mass of pentane}}{M_{C_{5}H_{12}}} = \frac{20.4\, \text{g}}{84.17\, \text{g/mol}} = 0.242\, \text{mol} $$

Apply stoichiometry to calculate the moles of water produced

From the balanced equation, we know that for the combustion of one mole of pentane, six moles of water are produced: $$ \mathrm{C}_{5} \mathrm{H}_{12}(l) \longrightarrow 6\mathrm{H}_{2} \mathrm{O}(l) $$ So, we can calculate the moles of water produced when 0.242 moles of pentane are burned: $$ \text{moles of water} = 0.242\, \text{mol} \times 6 = 1.452\, \text{mol} $$

Convert the moles of water to mass

Finally, we can calculate the mass of water produced using its molar mass (18.02 g/mol): $$ \text{mass of water} = \text{moles of water} \times M_{H_{2}O} = 1.452\, \text{mol} \times 18.02\, \text{g/mol} = 26.15\, \text{g} $$ Thus, 26.15 grams of water can be produced when 20.4 grams of pentane are burned in excess oxygen, assuming 100% yield.

Key concepts

Chemical Equation Balancing

Chemical equation balancing is a crucial step in understanding any chemical reaction, especially something as energetic as combustion. Balancing a chemical equation means ensuring that there are equal numbers of each type of atom on both sides of the equation. This is essential because atoms are neither created nor destroyed in chemical reactions; they merely rearrange.
To balance the combustion chemical equation for pentane, we start by writing down the formulas of the reactants and products:

• Pentane (\(\text{C}_5\text{H}_{12}\)) burns in oxygen (\(\text{O}_2\)).
• The products of complete combustion are carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)).

Next, you determine the correct coefficients to balance the equation. First, balance the carbon atoms. In \(\text{C}_5\text{H}_{12}\), there are 5 carbon atoms, so you put a coefficient of 5 in front of \(\text{CO}_2\). Then balance the hydrogen atoms: \(\text{C}_5\text{H}_{12}\) has 12 hydrogen atoms, so you'll need 6 \(\text{H}_2\text{O}\) molecules since each has 2 hydrogen atoms.
Lastly, balance the oxygen atoms. You will need 16 oxygen atoms from \(\text{O}_2\), equivalent to 8 \(\text{O}_2\) molecules. Thus, the balanced equation should read:\[\text{C}_5\text{H}_{12}(l) + 8\text{O}_2(g) \rightarrow 5\text{CO}_2(g) + 6\text{H}_2\text{O}(l)\]
Balancing equations helps in determining the proportions of reactants and products involved in chemical reactions.

Mole Calculations

Mole calculations allow us to convert between mass, moles, and molecules, making them pivotal in chemistry. To convert mass to moles, you need the molar mass—a measure of how much one mole of a substance weighs.
For pentane, we calculate its molar mass by adding the atomic masses of all atoms in the molecule. Pentane consists of five carbon atoms and twelve hydrogen atoms:

• Carbon has an atomic mass of 12.01 g/mol.
• Hydrogen has an atomic mass of 1.01 g/mol.

Thus, the molar mass of pentane is calculated as:\[M_{\text{C}_5\text{H}_{12}} = (5 \times 12.01 \,\text{g/mol}) + (12 \times 1.01 \,\text{g/mol}) = 84.17 \,\text{g/mol}\]
For 20.4 grams of pentane, the number of moles is found by dividing the mass by its molar mass:\[\text{moles of pentane} = \frac{20.4 \,\text{g}}{84.17 \,\text{g/mol}} = 0.242 \,\text{mol}\]
Using the balanced equation, we can relate this to the amount of water produced. If burning one mole of pentane produces 6 moles of water, burning 0.242 moles of pentane will result in:\[0.242 \,\text{mol} \times 6 = 1.452 \,\text{mol of } \text{H}_2\text{O}\]
Mole calculations are like a bridge connecting theoretical chemistry with practical measurements.

Combustion Reactions

Combustion reactions are fascinating and widely encountered in everyday life. These reactions are a kind of exothermic reaction, meaning they release energy, typically in the form of heat and light. During a combustion reaction, a substance combines with oxygen, resulting in the formation of oxides and the release of energy.
Let's talk about the combustion of pentane, a typical hydrocarbon combustion reaction:

• Fuel: Pentane (\(\text{C}_5\text{H}_{12}\)), composed of carbon and hydrogen.
• Oxidant: Oxygen (\(\text{O}_2\)), necessary for the reaction to occur.
• Products: Carbon dioxide (\(\text{CO}_2\)), and water (\(\text{H}_2\text{O}\)).

During the combustion, bonds between carbon and hydrogen in the fuel break, and new bonds form with oxygen, creating \(\text{CO}_2\) and \(\text{H}_2\text{O}\). This process also liberates significant energy, explaining why combustion is harnessed for heating and power.
In a typical balanced combustion equation, you can observe stoichiometry at play—each molecule of fuel needs a specific number of oxygen molecules to fully combust, turning into the corresponding amounts of carbon dioxide and water, all while obeying the law of conservation of mass.
Combustion reactions are pivotal not only in theoretical chemistry but also in real-world applications like engines and furnaces.