Question

Physicians who specialize in sports medicine routinely treat athletes and dancers. Ethyl chloride, a local anesthetic commonly used for simple injuries, is the product of the combination of ethylene with hydrogen chloride: $$ \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{HCl}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(g) $$ Assume that \(0.100 \mathrm{~kg}\) of \(\mathrm{C}_{2} \mathrm{H}_{4}\) and \(0.100 \mathrm{~kg}\) of \(\mathrm{HCl}\) react. (a) How many molecules of gas (reactants plus products) are present when the reaction is complete? (b) How many moles of gas are present when half the product forms?

Short answer

Total molecules (complete reaction) = \(5.45 \times 10^{24}\).Moles of gas (half product) = 3.564 moles.

Step-by-step solution

- Calculate the Moles of Each Reactant

First, determine the number of moles of each reactant using the given masses and molar masses.The molar mass of \(\text{C}_2\text{H}_4\) is \(\text{2} \times \text{12.01} + \text{4} \times \text{1.01} = 28.06 \text{ g/mol}\).The molar mass of \(\text{HCl}\) is \(\text{1.01} + \text{35.45} = 36.46 \text{ g/mol}\).Number of moles of \(\text{C}_2\text{H}_4\) = \(\frac{0.100 \text{ kg}}{28.06 \text{ g/mol}} \times 1000 \text{ g/kg} = 3.564 \text{ moles}\)Number of moles of \(\text{HCl}\) = \(\frac{0.100 \text{ kg}}{36.46 \text{ g/mol}} \times 1000 \text{ g/kg} = 2.742 \text{ moles}\)

- Identify the Limiting Reactant

Determine the limiting reactant by comparing the mole ratio of the reactants from the balanced equation.The balanced equation is \(\text{C}_2\text{H}_4 + \text{HCl} \to \text{C}_2\text{H}_5\text{Cl}\).The mole ratio is \(1:1\).Since there are fewer moles of \(\text{HCl}\), \(\text{HCl}\) is the limiting reactant.

- Calculate Moles of Product Formed

Using the limiting reactant, calculate the moles of \(\text{C}_2\text{H}_5\text{Cl}\) formed.Since \(\text{HCl}\) is the limiting reactant, it will determine the amount of product:Moles of \(\text{C}_2\text{H}_5\text{Cl}\) = Moles of \(\text{HCl}\) = \(\text{2.742 moles}\)

- Calculate Total Number of Gas Molecules (Part a)

Using Avogadro's number \(\text{6.022} \times 10^{23} \text{ molecules/mole}\), calculate the total number of molecules involved.Total moles of reactants = \(3.564 \text{ (C}_2\text{H}_4\text{)} + 2.742 \text{ (HCl)} = 6.306 \text{ moles}\)Total moles of products = \(2.742 \text{ moles (C}_2\text{H}_5\text{Cl)}\)Total moles of reactants and products = \(6.306 + 2.742 = 9.048 \text{ moles}\)Total molecules = \(9.048 \text{ moles} \times 6.022 \times 10^{23} \text{ molecules/mole} = 5.45 \times 10^{24} \text{ molecules}\)

- Calculate Moles of Gas When Half the Product Forms (Part b)

When half of the product forms, it implies half of the limiting reactant has reacted.Moles of \(\text{C}_2\text{H}_5\text{Cl}\) = \( \frac{2.742}{2} = 1.371 \text{ moles}\)Moles of \(\text{C}_2\text{H}_4\text{ remaining} = 3.564 - 1.371 = 2.193 \text{ moles}\)Moles of HCl remaining = 0Total moles of gas = Moles of \(\text{C}_2\text{H}_4\text{ remaining} + \text{HCl remaining} + \text{C}_2\text{H}_5\text{Cl}\text{ formed} = 2.193 + 0 + 1.371 = 3.564 \text{ moles}\)

Key concepts

limiting reactant

In any chemical reaction, the limiting reactant is the substance that runs out first and thus limits the amount of product that can be formed.
To find the limiting reactant, compare the mole ratio of the reactants given in the problem to the mole ratio in the balanced equation.
For example, in the reaction between ethylene \(\text{C}_2\text{H}_4\) and hydrogen chloride \(\text{HCl}\), the balanced equation is \(\text{C}_2\text{H}_4 + \text{HCl} \to \text{C}_2\text{H}_5\text{Cl}\) where the stoichiometric ratio is 1:1.
In the given exercise, if you have 3.564 moles of \(\text{C}_2\text{H}_4\) and 2.742 moles of \(\text{HCl}\), the \(\text{HCl}\) will be used up first. Hence, \(\text{HCl}\) is the limiting reactant.

moles of reactants and products

Understanding moles in chemical reactions is crucial.
A mole represents a quantity \(\text{6.022} \times 10^{23}\) of anything, typically atoms or molecules. In the given exercise, you convert the mass of each reactant to moles using their respective molar masses.
The reaction \(\text{C}_2\text{H}_4 + \text{HCl} \to \text{C}_2\text{H}_5\text{Cl}\) features the following calculations:

• Number of moles of \(\text{C}_2\text{H}_4\) = \(0.100 \text{ kg} \times \frac{1000 \text{ g/kg}}{28.06 \text{ g/mol}} = 3.564 \text{ moles}\)
• Number of moles of \(\text{HCl}\) = \(0.100 \text{ kg} \times \frac{1000 \text{ g/kg}}{36.46 \text{ g/mol}} = 2.742 \text{ moles}\)

The limiting reactant here is \(\text{HCl}\), meaning it will dictate the amount of product, which forms in a 1:1 molar ratio.

Avogadro's number

Avogadro's number \(6.022 \times 10^{23}\) is a key concept in chemistry, defining the number of particles (atoms, molecules, etc.) in a mole.
In the exercise, this number helps us convert moles to molecules.
After finding the total number of moles involved, you multiply by Avogadro's number to get total molecules.
For example, for a total of 9.048 moles of gas (reactants and products), the total number of molecules would be:
\(\text{9.048 moles} \times 6.022 \times 10^{23} \text{ molecules/mole} = 5.45 \times 10^{24} \text{ molecules}\).
This large number exemplifies how Avogadro's constant bridges the microscopic world of atoms to our macroscopic world.