Question

If \(1.5 \mathrm{~mol}\) of each of the following compounds is completely combusted in oxygen, which one will produce the largest number of moles of \(\mathrm{H}_{2} \mathrm{O}\) ? Which will produce the least? Explain. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}, \mathrm{C}_{3} \mathrm{H}_{\mathrm{B}}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{3}\)

Short answer

In the complete combustion of 1.5 moles of each compound, the least number of moles of H2O is produced by \(C_2H_5OH\) (4.5 moles), while both \(C_3H_8\) and \(CH_3CH_2COCH_3\) produce the most, with 6 moles of H2O each. This is determined by writing the balanced combustion reactions, calculating the moles of H2O produced by each compound, and comparing the results.

Step-by-step solution

1. Write the balanced combustion reactions for each compound

: For the given compounds, we need to find balanced combustion equations. The products of combustion reactions involving hydrocarbons are always carbon dioxide (CO2) and water (H2O). a) Combustion of \(C_2H_5OH\): \[C_2H_5OH + O_2 \rightarrow CO_2 + H_2O\] To balance the equation, we can use the following steps: - Balance the carbon atoms: There are 2 carbon atoms in \(C_2H_5OH\), so we need 2 \(CO_2\) molecules. - Balance the hydrogen atoms: There are 6 hydrogen atoms in \(C_2H_5OH\), so we need 3 \(H_2O\) molecules. - Balance the oxygen atoms: There are 7 oxygen atoms in total on the right side, so we need \(3.5~O_2\) molecules on the left side. So, the balanced equation is: \[C_2H_5OH + 3.5~O_2 \rightarrow 2~CO_2 + 3~H_2O\] b) Combustion of \(C_3H_8\): \[C_3H_8 + O_2 \rightarrow CO_2 + H_2O\] To balance the equation: - Balance the carbon atoms: There are 3 carbon atoms in \(C_3H_8\), so we need 3 \(CO_2\) molecules. - Balance the hydrogen atoms: There are 8 hydrogen atoms in \(C_3H_8\), so we need 4 \(H_2O\) molecules. - Balance the oxygen atoms: There are 10 oxygen atoms in total on the right side, so we need 5 \(O_2\) molecules on the left side. So, the balanced equation is: \[C_3H_8 + 5~O_2 \rightarrow 3~CO_2 + 4~H_2O\] c) Combustion of \(CH_3CH_2COCH_3\): \[CH_3CH_2COCH_3 + O_2 \rightarrow CO_2 + H_2O\] To balance the equation: - Balance the carbon atoms: There are 4 carbon atoms in \(CH_3CH_2COCH_3\), so we need 4 \(CO_2\) molecules. - Balance the hydrogen atoms: There are 8 hydrogen atoms in \(CH_3CH_2COCH_3\), so we need 4 \(H_2O\) molecules. - Balance the oxygen atoms: There are 9 oxygen atoms in total on the right side, so we need \(4.5~O_2\) molecules on the left side. So, the balanced equation is: \[CH_3CH_2COCH_3 + 4.5~O_2 \rightarrow 4~CO_2 + 4~H_2O\]

2. Determine the moles of H2O produced by 1.5 moles of each compound

: Next, we can use stoichiometry to find the moles of H2O produced by the combustion of 1.5 moles of each compound. a) Combustion of 1.5 moles \(C_2H_5OH\): According to the balanced equation, 1 mole of \(C_2H_5OH\) produces 3 moles of H2O. Therefore, 1.5 moles of \(C_2H_5OH\) will produce: \(1.5 \cdot 3 = 4.5\) moles of H2O b) Combustion of 1.5 moles \(C_3H_8\): According to the balanced equation, 1 mole of \(C_3H_8\) produces 4 moles of H2O. Therefore, 1.5 moles of \(C_3H_8\) will produce: \(1.5 \cdot 4 = 6\) moles of H2O c) Combustion of 1.5 moles \(CH_3CH_2COCH_3\): According to the balanced equation, 1 mole of \(CH_3CH_2COCH_3\) produces 4 moles of H2O. Therefore, 1.5 moles of \(CH_3CH_2COCH_3\) will produce: \(1.5 \cdot 4 = 6\) moles of H2O

3. Compare the moles of H2O produced and determine the compound producing the most and the least H2O

: Comparing the moles of H2O formed from the combustion of each compound: - \(C_2H_5OH\) produces 4.5 moles of H2O - \(C_3H_8\) produces 6 moles of H2O - \(CH_3CH_2COCH_3\) produces 6 moles of H2O From the comparison, it is clear that the combustion of \(C_2H_5OH\) produces the least number of moles of H2O, while both \(C_3H_8\) and \(CH_3CH_2COCH_3\) produce the most (largest) number of moles of H2O.

Key concepts

Balanced Chemical Equations

Writing balanced chemical equations is a crucial skill in chemistry. It ensures that the number of each type of atom is the same on both sides of the equation, conserving the mass in chemical reactions. In combustion reactions, hydrocarbons combine with oxygen to form carbon dioxide and water. For example, ethyl alcohol (\(C_2H_5OH\)) reacts with oxygen to form:
\[C_2H_5OH + 3.5~O_2 \rightarrow 2~CO_2 + 3~H_2O\] This equation shows that:

• 2 carbon atoms in ethanol lead to 2 carbon dioxide molecules.
• 6 hydrogen atoms lead to 3 water molecules.
• Oxygen is adjusted with fractional coefficients.

Balancing equations involves adjusting the coefficients (numbers before molecules) to ensure equal atoms on each side. It's like a recipe; each ingredient (atom) should be used fully without leftovers.

Stoichiometry

Stoichiometry is all about measuring and calculating the amounts of reactants and products in a chemical reaction. It involves converting between moles, mass, and number of molecules using balanced equations. In our exercise, stoichiometry tells us how many moles of water are produced when 1.5 moles of each compound are combusted.

• For \(C_2H_5OH\), each mole produces 3 moles of \(H_2O\), so 1.5 moles yield \(1.5 \times 3 = 4.5\) moles.
• For \(C_3H_8\), each mole produces 4 moles of \(H_2O\), resulting in \(1.5 \times 4 = 6\) moles.
• Similarly, \(CH_3CH_2COCH_3\) also gives \(6\) moles.

By relating moles through these calculations, stoichiometry provides a quantitative insight into chemical processes.

Hydrocarbon Combustion

Hydrocarbon combustion is a chemical reaction where a hydrocarbon reacts with oxygen to produce carbon dioxide and water. These reactions are fundamental to energy production in our world.

• Typical hydrocarbons include ethane (\(C_2H_5OH\)), propane (\(C_3H_8\)), and butanone (\(CH_3CH_2COCH_3\)).
• The balanced equations help us understand the exact requirements for oxygen and yield of \(CO_2\) and \(H_2O\).

Understanding the combustion of hydrocarbons enables us to calculate emissions, energy content, and efficiency of fuels. These reactions are key in many everyday processes, from heating to running engines.