Question

If \(1.5 \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}, 1.5 \mathrm{mol} \mathrm{C}_{3} \mathrm{H}_{8},\) and 1.5 \(\mathrm{mol} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{3}\) are completely combusted in oxygen, which produces the largest number of moles of \(\mathrm{H}_{2} \mathrm{O} ?\) Which produces the least? Explain.

Short answer

Propane (\(C_{3}H_{8}\)) and methyl ethyl ketone (\(CH_{3}CH_{2}COCH_{3}\)) produce the largest number of moles of water (\(H_{2}O\)) with a 1:4 mole ratio, while ethanol (\(C_{2}H_{5}OH\)) produces the least with a 1:3 mole ratio when completely combusted in oxygen.

Step-by-step solution

Combustion reaction of C2H5OH

The balanced combustion reaction of ethanol (\(C_{2}H_{5}OH\)) is given by: \[C_{2}H_{5}OH + 3O_{2} \rightarrow 2CO_{2} + 3H_{2}O\] 2.

Combustion reaction of C3H8

The balanced combustion reaction of propane (\(C_{3}H_{8}\)) is given by: \[C_{3}H_{8} + 5O_{2} \rightarrow 3CO_{2} + 4H_{2}O\] 3.

Combustion reaction of CH3CH2COCH3

The balanced combustion reaction of methyl ethyl ketone (\(CH_{3}CH_{2}COCH_{3}\)) is given by: \[CH_{3}CH_{2}COCH_{3} + \frac{11}{2}O_{2} \rightarrow 3CO_{2} + 4H_{2}O\] 4.

Comparing mole ratios for each compound

#_Ethanol:_ In the combustion reaction of ethanol, each mole of ethanol produces 3 moles of water. The mole ratio is \(1:3\). _Propane:_ In the combustion reaction of propane, each mole of propane produces 4 moles of water. The mole ratio is \(1:4\). _Methyl ethyl ketone:_ In the combustion reaction of methyl ethyl ketone, each mole of the compound produces 4 moles of water. The mole ratio is \(1:4\). 5.

Determining which compound produces the most and least amount of water

Both propane and methyl ethyl ketone have a mole ratio of \(1:4\) (one mole of each compound producing 4 moles of water), while ethanol has a mole ratio of \(1:3\) (one mole of ethanol producing 3 moles of water). Therefore, propane and methyl ethyl ketone produce the largest number of moles of water, while ethanol produces the least.

Key concepts

Ethanol Combustion

Ethanol combustion is a type of chemical reaction where ethanol ( \(C_2H_5OH\)) reacts with oxygen to produce carbon dioxide and water. In a balanced ethanol combustion equation, \[C_{2}H_{5}OH + 3O_{2} \rightarrow 2CO_{2} + 3H_{2}O\]each mole of ethanol produces three moles of water.
Combustion reactions work by breaking the bonds of ethanol and oxygen, then forming new bonds to create water and carbon dioxide.

• Ethanol has a structure of two carbon atoms and an OH group, making it a simple alcohol.
• It is widely recognized for its use as a biofuel and in alcoholic beverages.
• In terms of energy production, ethanol releases energy but forms fewer moles of water compared to some other fuels like propane and methyl ethyl ketone.

Understanding ethanol combustion helps in calculating the efficiency of biofuels and comparing their outputs to other fuels. For instance, the mole ratio indicates that for every mole of ethanol, three moles of water are produced.

Propane Combustion

Propane is another common fuel characterized by the chemical formula \(C_3H_8\). Its combustion can be represented by the balanced equation:\[C_{3}H_{8} + 5O_{2} \rightarrow 3CO_{2} + 4H_{2}O\]A combustion reaction involving propane produces four moles of water for every mole of propane.

• Propane is known for being a clean-burning fuel, often used in heating, cooking, and in some vehicles.
• The structure comprises of three carbon atoms and eight hydrogen atoms, which contributes to its energy-rich nature.
• It is stored as a liquid in portable tanks and is popular in areas without natural gas connections.

Calculating the products of propane combustion involves setting up a stoichiometric calculation between the reactants and products, allowing for an understanding of the efficiency and comparison with other fuels. With a higher mole ratio than ethanol, propane provides a greater yield of water, showcasing its potential as an energy source.

Methyl Ethyl Ketone Combustion

Methyl ethyl ketone (MEK), with the formula \(CH_3CH_2COCH_3\), combusts in a reaction involving oxygen to produce carbon dioxide and water. The balanced chemical equation is:\[CH_{3}CH_{2}COCH_{3} + \frac{11}{2}O_{2} \rightarrow 3CO_{2} + 4H_{2}O\]This reaction shows that one mole of methyl ethyl ketone yields four moles of water.

• MEK is a solvent with applications in processes such as wax removal and coatings.
• Its combustion indicates it as a significant source of energy with water and carbon dioxide as byproducts.
• Typically, it's less common as a fuel source because it's primarily used industrially.

Studying methyl ethyl ketone combustion can illuminate efficiency traits similar to propane, as both produce four moles of water per mole of reactant. Comparing MEK with ethanol and propane underlines its industrial importance and potential energy production capabilities.