Question

DDT, an insecticide harmful to fish, birds, and humans, is produced by the following reaction: $$2{\mathrm{C}}_{\mathrm{f}}{\mathrm{H}}_5\mathrm{Cl}+{\mathrm{C}}_2{\mathrm{HOCl}}_3⟶{\mathrm{C}}_{14}{\mathrm{H}}_9{\mathrm{Cl}}_5+{\mathrm{H}}_2\mathrm{O}$$ In a government lab, 1142 g of chlorobenzene is reacted with 485 g of chloral. a. What mass of DDT is formed, assuming 100$\mathrm{\%}$ yield? b. Which reactant is limiting? Which is in excess? c. What mass of the excess reactant is left over? d. If the actual yield of DDT is 200.0 g, what is the percent yield?

Short answer

a. 1032.52 g of DDT is formed, assuming a 100% yield. b. Chloral is the limiting reactant, and chlorobenzene is in excess. c. 486.31 g of excess chlorobenzene is left over. d. The percent yield of DDT is 19.37%.

Step-by-step solution

1. Calculate the moles of reactants:

First, calculate the number of moles for each reactant by dividing the given mass by its molar mass. The molar masses can be determined from the atomic weights in the periodic table: Chlorobenzene ($C_6{H}_{5}Cl$): $12.01(6)+1.01(5)+35.45=112.56g/mol$ Chloral ($C_{2}HOC{l}_3$): $12.01(2)+1.01+16.00+35.45(3)=166.43g/mol$ Moles of chlorobenzene: $\frac{1142}{112.56}=10.146mol$ Moles of chloral: $\frac{485}{166.43}=2.913mol$

2. Find the limiting reactant and the mass of DDT formed:

To find the limiting reactant, compare the mole ratio between reactants by dividing the moles calculated by their respective stoichiometric coefficients from the balanced equation: Chlorobenzene: $\frac{10.146}{2}=5.073$ Chloral: $\frac{2.913}{1}=2.913$ Since the value for chloral is lower, it is the limiting reactant. Now we can calculate the mass of DDT produced using the stoichiometry of the balanced equation. The molar mass of DDT ($C_{14}{H}_{9}C{l}_5$) is: $12.01(14)+1.01(9)+35.45(5)=354.44g/mol$. The equation tells us that 1 mole of chloral produces 1 mole of DDT. Therefore, using the moles of the limiting reactant, chloral: Mass of DDT = moles of chloral × molar mass of DDT = $2.913\ast 354.44=1032.52g$ However, the problem states that we assume a 100% yield, so we can simply calculate the mass directly: Mass of DDT formed: $1032.52g$

3. Calculate the leftover mass of the excess reactant:

Since chlorobenzene is the excess reactant, we need to determine the amount that reacted with 2.913 moles of chloral. From the balanced equation, 2 moles of chlorobenzene react with 1 mole of chloral: Moles of chlorobenzene reacted: $(2.913\ast 2)=5.826mol$ Now convert the moles of reacted chlorobenzene to mass: Mass of reacted chlorobenzene: $5.826\ast 112.56=655.69g$ Subtract this mass from the initial mass to find the leftover mass: Leftover mass of chlorobenzene: $1142-655.69=486.31g$

4. Calculate the percent yield:

The actual yield of DDT is given as 200.0 g. To find the percent yield, we can use the following formula: Percent yield = $\frac{Actualyield}{Theoreticalyield}\ast 100\mathrm{\%}$ Percent yield = $\frac{200.0}{1032.52}\ast 100\mathrm{\%}=19.37\mathrm{\%}$ So, to answer each part of the problem: a. The mass of DDT formed is 1032.52 g, assuming a 100% yield. b. The limiting reactant is chloral, and the excess reactant is chlorobenzene. c. 486.31 g of chlorobenzene remains. d. The percent yield of DDT is 19.37%.

Key concepts

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that is completely consumed first. Once this reactant is used up, the reaction cannot proceed further, and it limits the amount of product that can be formed. Identifying the limiting reactant is crucial in stoichiometry because it determines the maximum amount of product possible from the given reactants.

To find the limiting reactant, first, we calculate the moles of each reactant using their masses and molar masses. Then, we use the balanced chemical equation to understand the mole ratio between them. By comparing these moles adjusted to their stoichiometric coefficients, we can identify which reactant is limiting. In the given problem, chloral (${\mathrm{C}}_2{\mathrm{HOCl}}_3$) is the limiting reactant because its available mole ratio becomes lower compared to chlorobenzene.

Understanding the concept of the limiting reactant helps in various applications like optimizing reactions and determining the cost-effectiveness of industrial processes.

Percent Yield

The percent yield of a reaction expresses the efficiency of the reaction, comparing the actual yield to the theoretical yield. The theoretical yield is the maximum amount of product that can be produced based on the limiting reactant, assuming perfect efficiency and no losses. The actual yield, however, is what is obtained practically, often less due to incomplete reactions, side reactions, or losses during processing.

The formula for percent yield is:

• Percent Yield = $\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\mathrm{\%}$

In our exercise, the actual yield of DDT produced was 200.0 g, while the theoretical yield was calculated to be 1032.52 g. Thus, the percent yield was:

• Percent Yield = $\frac{200.0}{1032.52}\times 100\mathrm{\%}=19.37\mathrm{\%}$

Having a percent yield allows chemists to assess how well a reaction has performed and highlight areas to improve efficiency.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They provide the necessary details for understanding how reactants transform into products under specific conditions. In a balanced chemical equation, the numbers of atoms of each element are equal on both sides of the equation. This balance obeys the conservation of mass principle, ensuring that no atoms are lost or created in the reaction.

In our case, the equation for producing DDT is given as:

• $2{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{Cl}+{\mathrm{C}}_2{\mathrm{HOCl}}_3\to {\mathrm{C}}_{14}{\mathrm{H}}_9{\mathrm{Cl}}_5+{\mathrm{H}}_2\mathrm{O}$

This equation tells us that two moles of chlorobenzene react with one mole of chloral to produce one mole of DDT and one mole of water. Understanding chemical equations is essential for predicting the amounts of products and reactants involved in reactions, aiding in calculations such as determining the limiting reactant, theoretical yields, and more.

By mastering chemical equations, students can better grasp the changes taking place during reactions, making stoichiometry calculations more intuitive and accurate.