Question

When iron(III) chloride \(\left(\mathrm{FeCl}_{3}\right)\) reacts in an atmosphere of pure oxygen, the following occurs: $$4 \mathrm{FeCl}_{3}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{gv}) \rightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+6 \mathrm{Cl}_{2}(\mathrm{g})$$ If 45.0 \(\mathrm{g}\) of \(\mathrm{FeCl}_{3}\) reacts and 20.5 \(\mathrm{g}\) of iron(III) oxide is recovered, determine the percent yield. (Chapter 11\()\)

Short answer

The percent yield for the reaction of 45.0 g of \(\mathrm{FeCl}_{3}\) producing 20.5 g of iron(III) oxide is 92.6%.

Step-by-step solution

Calculate the moles of the reactant

First, we need to calculate the amount of moles of \(\mathrm{FeCl}_{3}\). To do so, divide the mass of the substance by its molar mass. The molar mass of \(\mathrm{FeCl}_{3}\) is \((1 \times 55.85) + (3 \times 35.45) = 162.20\,\mathrm{g/mol}\). Calculate the moles by using the mass given in the problem (45.0 g): $$\text{moles of } \mathrm{FeCl}_{3} = \frac{45.0\,\mathrm{g}}{162.20\,\mathrm{g/mol}} = 0.27729\,\mathrm{mol}$$

Calculate the theoretical yield using stoichiometry

Now, refer to the balanced chemical equation and use the stoichiometry ratios to calculate the moles of iron(III) oxide which can be produced by 0.27729 mol of \(\mathrm{FeCl}_{3}\): $$4 \mathrm{FeCl}_{3} \rightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}$$ Using the stoichiometric ratio: $$\text{moles of } \mathrm{Fe}_{2} \mathrm{O}_{3} = 0.27729\,\mathrm{mol} \times \frac{2\,\mathrm{mol}\, \mathrm{Fe}_{2} \mathrm{O}_{3} }{4\,\mathrm{mol}\, \mathrm{FeCl}_{3}} = 0.13864\,\mathrm{mol}$$ Now, convert the moles of iron(III) oxide to grams using the molar mass of iron(III) oxide, which is \((2 \times 55.85) + (3 \times 16) = 159.7\,\mathrm{g/mol}\): $$\text{theoretical yield } = 0.13864\,\mathrm{mol} \times 159.7\,\mathrm{g/mol} = 22.12\,\mathrm{g}$$

Calculate the percent yield

Now, compare the theoretical yield (22.12 g) with the actual yield (20.5 g) to find the percent yield: $$\text{percent yield } = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$$ $$\text{percent yield } = \frac{20.5\,\mathrm{g}}{22.12\,\mathrm{g}} \times 100 = 92.6\%$$

Key concepts

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships of reactants and products in a chemical reaction. It is based on the law of conservation of mass, which states that in a chemical reaction, matter is neither created nor destroyed. In practice, stoichiometry involves using a balanced chemical equation to calculate the amounts of reactants needed or the amounts of products formed.

For instance, if we know the quantity of one reactant, we can use stoichiometric ratios from the balanced equation to predict the amount of product that can be formed if the reaction goes to completion. To improve your mastery of stoichiometry, focus on accurately balancing chemical equations and understanding the mole concept, which enables conversion between mass, number of particles, and volume in gases at standard conditions.

Theoretical Yield

Theoretical yield refers to the maximum amount of product that can be generated from a given amount of reactants under ideal conditions. This is a key concept in stoichiometry and helps chemists estimate how much product should be expected from a reaction. To calculate theoretical yield, one first determines the limiting reactant, the substance that will be entirely consumed when the reaction goes to completion.

After identifying the limiting reactant and its quantity, we use the stoichiometric relationship of the balanced chemical equation to calculate the maximum possible amount of product formed. It is essential to consider this as it sets the standard for analyzing the efficiency of the reaction. Utilizing theoretical yield calculations, we strive to understand and predict the efficiency of chemical processes.

Chemical Reaction

A chemical reaction involves the transformation of one or more substances into different substances. This change occurs through the breaking and forming of chemical bonds. Reactions are represented by chemical equations that show the reactants on the left side and the products on the right.

To enhance understanding, it's crucial to recognize the pattern of a reaction, whether it's a combination, decomposition, single-replacement, or double-replacement reaction. In our iron(III) chloride example, oxygen reacts with iron(III) chloride in a combination reaction to yield iron(III) oxide and chlorine gas. Detailed knowledge of different types of reactions aids in predicting products and understanding the stoichiometric relationships between reactants and products.

Molar Mass

Molar mass is a fundamental concept in chemistry defined as the mass of one mole of a substance. It is expressed in grams per mole (g/mol) and can be calculated by summing the atomic masses of all atoms in a molecule, as determined by the periodic table.

Knowing the molar mass of compounds allows us to convert between the mass of a substance and the number of moles, laying the groundwork for stoichiometric calculations. This conversion is vital for all aspects of stoichiometry, as it allows chemists to measure substances on a molecular level and to calculate reactants or products in a chemical reaction. Keep in mind, accurate determination of molar masses from the molecular formula is a useful skill for any student studying chemistry.