Question

The combustion of one mole of liquid ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) , produces 1367 \(\mathrm{kJ}\) of heat. Calculate how much heat is produced when 235.0 \(\mathrm{g}\) of ethanol are combusted.

Short answer

When 235.0 g of ethanol is combusted, 6971.7 kJ of heat is produced.

Step-by-step solution

Find the molar mass of ethanol

First, we need to find the molar mass of ethanol, CH3CH2OH. The molar mass of carbon is 12.01 g/mol, hydrogen is 1.01 g/mol, and oxygen is 16.00 g/mol. There are 2 carbon atoms, 6 hydrogen atoms, and 1 oxygen atom in ethanol. Molar mass of ethanol = ( 2 * 12.01 g/mol) + (6 * 1.01 g/mol) + 16.00 g/mol = 46.07 g/mol

Convert mass of ethanol to moles

Now that we have the molar mass of ethanol, we can convert the given mass of ethanol in grams to moles using the molar mass as a conversion factor. Moles of ethanol = \(\frac{235.0 \,\text{g}}{46.07 \,\text{g/mol}} \)= 5.10 moles

Calculate the heat produced

We know that the combustion of 1 mole of ethanol produces 1367 kJ of heat. Therefore, we can calculate the heat produced when 5.10 moles of ethanol are combusted by multiplying the moles of ethanol by the heat released per mole. Heat produced = 5.10 moles * 1367 kJ/mol = 6971.7 kJ When 235.0 g of ethanol is combusted, 6971.7 kJ of heat is produced.

Key concepts

Understanding Molar Mass Calculation

The molar mass of a substance is simply the mass of one mole of its entities (atoms, molecules, etc.). It is calculated by summing the atomic masses of all the atoms present in a molecule. For ethanol, which has the molecular formula \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\), we need the atomic masses of carbon, hydrogen, and oxygen.
The atomic mass of carbon is 12.01 g/mol, hydrogen is 1.01 g/mol, and oxygen is 16.00 g/mol. Ethanol consists of two carbon atoms, six hydrogen atoms, and one oxygen atom. To find the molar mass, we multiply the number of each type of atom by their respective atomic masses, and then add those values together:

• Carbon: \(2 \times 12.01 \, \text{g/mol} = 24.02 \, \text{g/mol}\)
• Hydrogen: \(6 \times 1.01 \, \text{g/mol} = 6.06 \, \text{g/mol}\)
• Oxygen: \(1 \times 16.00 \, \text{g/mol} = 16.00 \, \text{g/mol}\)

Adding these up gives you the molar mass of ethanol, which is 46.07 g/mol.

Exploring Stoichiometry in Chemical Reactions

Stoichiometry is the aspect of chemistry that involves calculating the quantitative relationships of reactants and products in a chemical reaction. It allows us to determine how much of a substance is consumed or produced.
In the context of the combustion of ethanol, we want to find out how many moles of ethanol are involved. Since the problem states that 235.0 g of ethanol are combusted, and we've calculated the molar mass to be 46.07 g/mol, conversion is straightforward:

We use the formula: \[\text{Moles of ethanol} = \frac{\text{Mass of ethanol}}{\text{Molar mass of ethanol}}\]
Substituting the values in: \[\frac{235.0 \, \text{g}}{46.07 \, \text{g/mol}} = 5.10 \, \text{moles}\]
Doing stoichiometric calculations ensures correct proportions of reactants and products, which is crucial in practical applications like energy production.

Understanding Energy Conversion in Combustion

Energy conversion during a chemical reaction such as combustion involves transforming chemical energy stored in bonds to heat or other forms of energy. For ethanol, the combustion process releases heat as ethanol reacts with oxygen to form carbon dioxide and water.

Combustion of ethanol is an exothermic reaction, meaning it releases energy. As specified, the combustion of one mole of ethanol releases 1367 kJ. Thus, knowing the number of moles combusted, we multiply by the energy per mole to calculate total energy released.
For 5.10 moles of ethanol combusted, the heat produced is:\[5.10 \, \text{moles} \times 1367 \, \frac{\text{kJ}}{\text{mole}} = 6971.7 \, \text{kJ}\]
This process highlights the principles of energy conversion and conservation, providing insight into how chemical reactions can be utilized to produce energy for practical uses, such as in heating and powering engines.