Question

The reaction between \(113.4 \mathrm{~g}\) of \(\mathrm{I}_{2} \mathrm{O}_{5}\) and \(132.2 \mathrm{~g}\) of \(\mathrm{Br} \mathrm{F}_{3}\) was found to produce \(97.0 \mathrm{~g}\) of \(\mathrm{IF}_{5}\). The equation for the reaction is \(6 \mathrm{I}_{2} \mathrm{O}_{3}+20 \mathrm{BrF}_{3} \rightarrow 12 \mathrm{IF}_{5}+15 \mathrm{O}_{2}+10 \mathrm{Br}_{2}\) What is the per cent yield of IF \(_{5}\) ? a. \(37.7 \%\) b. \(75.4 \%\) c. \(150.8 \%\) d. \(57.4 \%\)

Short answer

The percent yield of IF₅ is 75.4% (option b).

Step-by-step solution

Calculate the Molar Mass

Determine the molar masses of the reactants and products. - Molar mass of \( \mathrm{I}_2\mathrm{O}_5 \) is approximately \( 333.8 \text{ g/mol} \).- Molar mass of \( \mathrm{BrF}_3 \) is approximately \( 136.9 \text{ g/mol} \).- Molar mass of \( \mathrm{IF}_5 \) is approximately \( 221.9 \text{ g/mol} \).

Convert Mass to Moles for Reactants

Calculate the moles of \( \mathrm{I}_2\mathrm{O}_5 \) and \( \mathrm{BrF}_3 \) using their molar masses.\( \text{Moles of } \mathrm{I}_2\mathrm{O}_5 = \frac{113.4 \text{ g}}{333.8 \text{ g/mol}} \approx 0.340 \text{ mol} \)\( \text{Moles of } \mathrm{BrF}_3 = \frac{132.2 \text{ g}}{136.9 \text{ g/mol}} \approx 0.966 \text{ mol} \)

Identify the Limiting Reactant

The balanced equation shows that 6 moles of \( \mathrm{I}_2\mathrm{O}_5 \) react with 20 moles of \( \mathrm{BrF}_3 \).Calculate the moles of \( \mathrm{BrF}_3 \) needed for 0.340 moles of \( \mathrm{I}_2\mathrm{O}_5 \):\( \frac{20}{6} \times 0.340 \approx 1.133 \text{ mol of } \mathrm{BrF}_3 \)Since only 0.966 moles of \( \mathrm{BrF}_3 \) are available, \( \mathrm{BrF}_3 \) is the limiting reactant.

Calculate Theoretical Yield of IF5

According to the reaction, 20 moles of \( \mathrm{BrF}_3 \) produce 12 moles of \( \mathrm{IF}_5 \). Therefore:\( \mathrm{Moles \ of \ IF}_5 \) produced from 0.966 moles of \( \mathrm{BrF}_3 \):\( \frac{12}{20} \times 0.966 \approx 0.579 \text{ mol of } \mathrm{IF}_5 \)Convert moles to grams:\( 0.579 \text{ mol} \times 221.9 \text{ g/mol} \approx 128.5 \text{ g of } \mathrm{IF}_5 \)

Calculate Percent Yield

Percent yield is calculated using the formula:\[ \text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100 \]Substituting the given values:\[ \text{Percent Yield} = \left(\frac{97.0 \text{ g}}{128.5 \text{ g}}\right) \times 100 \approx 75.4\% \]

Key concepts

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that is completely consumed first and thus determines the maximum amount of product that can be formed. This concept is essential because it affects the calculations for determining the theoretical yield of the reaction. Understanding which reactant "runs out" first can help predict the amounts of products formed.
To identify the limiting reactant, use the balanced chemical equation to compare the mole ratio between the reactants. For instance, in the given reaction between \(\mathrm{I}_2\mathrm{O}_5\) and \(\mathrm{BrF}_3\), the chemical equation indicates that 6 moles of \(\mathrm{I}_2\mathrm{O}_5\) react with 20 moles of \(\mathrm{BrF}_3\). Use these ratios to decide which reactant is limiting. If you calculate and find that you have insufficient \(\mathrm{BrF}_3\) to react with all the \(\mathrm{I}_2\mathrm{O}_5\), then \(\mathrm{BrF}_3\) is the limiting reactant.
Knowing the limiting reactant is crucial for calculating both the theoretical yield and percent yield of the reaction, which are fundamental considerations in industrial chemical processes and laboratory experiments alike.

Molar Mass

The molar mass of a compound is the mass of one mole of its molecules. It's calculated by summing the atomic masses of all the atoms that make up the molecule. The concept of molar mass is pivotal because it's used to convert between the mass of a substance and the moles of the substance, an essential step in stoichiometry.
Let's consider our example: the molar mass of \(\mathrm{I}_2\mathrm{O}_5\) is approximately \(333.8\,\text{g/mol}\), which means one mole of \(\mathrm{I}_2\mathrm{O}_5\) weighs 333.8 grams. Similarly, for \(\mathrm{BrF}_3\), the molar mass is approximately \(136.9\,\text{g/mol}\), and for the product \(\mathrm{IF}_5\), it is \(221.9\,\text{g/mol}\).
Understanding and calculating molar masses accurately enable you to determine how much of each reactant is involved in the reaction and how much product can be potentially formed. Always remember, the correct molar mass calculations are the bedrock of solving chemical stoichiometry problems.

Stoichiometry

Stoichiometry is the part of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It involves using a balanced equation to derive conversion factors that make mole-to-mole, mole-to-mass, and mass-to-mass conversions feasible. This concept allows chemists to predict the amount of product formed from given reactants.
For example, in the reaction of \(\mathrm{I}_2\mathrm{O}_5\) with \(\mathrm{BrF}_3\), the balanced equation tells us that 20 moles of \(\mathrm{BrF}_3\) can produce 12 moles of \(\mathrm{IF}_5\). This relationship helps in calculating the theoretical yield, where the stoichiometric coefficients (from the balanced chemical equation) are used to derive the proportion of reactants to products.
Stoichiometry is not just about making predictions; it's a critical tool in ensuring that chemical reactions proceed with maximum efficiency. Understanding stoichiometry ensures that you can adjust reactant quantities to minimize waste, calculate yields accurately, and thereby optimize chemical processes efficiently.