Question

Ethylene gas, \(\mathrm{C}_{2} \mathrm{H}_{4}\), reacts with water at high temperature to yield ethyl alcohol, \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}\). (a) How many grams of ethylene are needed to react with \(0.133\) mol of \(\mathrm{H}_{2} \mathrm{O}\) ? How many grams of ethyl alcohol will result? (b) How many grams of water are needed to react with \(0.371 \mathrm{~mol}\) of ethylene? How many grams of ethyl alcohol will result?

Short answer

(a) 3.73 g of ethylene; 6.13 g of ethyl alcohol. (b) 6.68 g of water; 17.09 g of ethyl alcohol.

Step-by-step solution

Write the Balanced Chemical Equation

The reaction between ethylene, \(\text{C}_2\text{H}_4\), and water \(\text{H}_2\text{O}\) to produce ethyl alcohol \(\text{C}_2\text{H}_6\text{O}\) is as follows:\[ \text{C}_2 \text{H}_4 (g) + \text{H}_2 \text{O} (l) \rightarrow \text{C}_2 \text{H}_6 \text{O} (l) \]This shows a 1:1:1 molar relationship among ethylene, water, and ethyl alcohol.

Calculate Molar Masses

Calculate the molar masses of the substances involved:- \(\text{C}_2\text{H}_4\): \(2 \times 12.01 + 4 \times 1.01 = 28.06\) g/mol.- \(\text{H}_2\text{O}\): \(2 \times 1.01 + 16.00 = 18.02\) g/mol.- \(\text{C}_2\text{H}_6\text{O}\): \(2 \times 12.01 + 6 \times 1.01 + 16.00 = 46.08\) g/mol.

Determine Ethylene Required for Given Water (Part a)

For \(0.133\) mol of \(\text{H}_2\text{O}\), the same moles of \(\text{C}_2\text{H}_4\) are needed:\[ 0.133 \text{ mol C}_2\text{H}_4 \times 28.06 \text{ g/mol} = 3.73 \text{ g C}_2\text{H}_4 \]

Calculate Ethyl Alcohol Produced (Part a)

Using the stoichiometry from the balanced equation, \(0.133\) mol of \(\text{C}_2\text{H}_6\text{O}\) is produced:\[ 0.133 \text{ mol C}_2\text{H}_6\text{O} \times 46.08 \text{ g/mol} = 6.13 \text{ g C}_2\text{H}_6\text{O} \]

Determine Water Required for Given Ethylene (Part b)

For \(0.371\) mol of \(\text{C}_2\text{H}_4\), the same moles of \(\text{H}_2\text{O}\) are needed:\[ 0.371 \text{ mol H}_2\text{O} \times 18.02 \text{ g/mol} = 6.68 \text{ g H}_2\text{O} \]

Calculate Ethyl Alcohol Produced (Part b)

Using the stoichiometry of the balanced equation for \(0.371\) mol of \(\text{C}_2\text{H}_6\text{O}\):\[ 0.371 \text{ mol C}_2\text{H}_6\text{O} \times 46.08 \text{ g/mol} = 17.09 \text{ g C}_2\text{H}_6\text{O} \]

Key concepts

Molar Mass Calculation

Calculating the molar mass is a crucial skill in chemical stoichiometry. The molar mass of a compound refers to the mass of one mole of that substance, expressed in grams per mole (g/mol). To find the molar mass, you sum the atomic masses of all the atoms in the compound. Let's look deeper into this process using the compounds involved in the reaction of ethylene with water.

• For ethylene (\(\text{C}_2\text{H}_4\)), calculate its molar mass by adding the atomic masses of carbon and hydrogen. Carbon has an atomic mass of 12.01 g/mol, and hydrogen is 1.01 g/mol. Therefore, the molar mass is calculated as: \(2 \times 12.01 + 4 \times 1.01 = 28.06\ \text{g/mol}\)
• Water (\(\text{H}_2\text{O}\)) has a molar mass of: \(2 \times 1.01 + 16.00 = 18.02\ \text{g/mol}\)
• Ethyl alcohol (\(\text{C}_2\text{H}_6\text{O}\)) involves a bit more calculation as it contains carbon, hydrogen, and oxygen: \(2 \times 12.01 + 6 \times 1.01 + 16.00 = 46.08\ \text{g/mol}\)

Calculating molar mass correctly helps in converting between moles and grams, which is essential for solving stoichiometry problems.

Balanced Chemical Equation

A balanced chemical equation is your blueprint for understanding a chemical reaction's stoichiometry. It shows exactly how atoms are conserved in the reaction, following the Law of Conservation of Mass.
The balanced equation for the reaction of ethylene with water to form ethyl alcohol is:\[ \text{C}_2 \text{H}_4 (g) + \text{H}_2 \text{O} (l) \rightarrow \text{C}_2 \text{H}_6 \text{O} (l) \]This equation tells us that:

• One mole of ethylene reacts with one mole of water.
• The product, ethyl alcohol, is also in a 1:1 molar ratio with the reactants.

This simple 1:1:1 ratio makes calculations straightforward because it implies that the moles of reactants used will equal the moles of products made, simplifying the computation of mass and moles in any stoichiometric calculations.

Chemical Reaction

Understanding a chemical reaction involves knowing the substances transforming and the products formed. A chemical reaction involves the breaking of old bonds and forming new ones, while rearranging atoms between different states or compounds.
In our exercise, the reaction involves ethylene (\(\text{C}_2\text{H}_4\)) and water (\(\text{H}_2\text{O}\)) reacting under high temperature conditions.

• Ethylene is converted to ethyl alcohol (\(\text{C}_2\text{H}_6\text{O}\)).
• This reaction involves using water as a reactant alongside ethylene.
• Conditions such as high temperature may be necessary to make sure the reaction proceeds effectively.

The entire transformation can be summed up by the balanced chemical equation previously discussed, which shows the specific reactants and products in precise quantities required and produced. Understanding such reactions is fundamental in predicting the behavior and yield of reactions in chemical processes.