Problem 141 - Introductory Chemistry Atoms First

Question

Consider the unbalanced chemical equation \(\mathrm{Cl}_{2} \mathrm{O}_{7}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{HClO}_{4}\) The reaction is carried out at \(82.0 \%\) yield and gives \(52.8 \mathrm{~g}\) of \(\mathrm{HClO}_{4}\) (a) What is the theoretical yield of \(\mathrm{HClO}_{4}\) ? (b) How many grams of \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) and \(\mathrm{H}_{2} \mathrm{O}\) were consumed in the reaction?

Step-by-Step Solution

Answer

The theoretical yield of \(\mathrm{HClO}_{4}\) is approximately \(64.4 \mathrm{~g}\). In the reaction, about \(58.6\,\mathrm{g}\) of \(\mathrm{Cl}_2\mathrm{O}_7\) and \(11.5\,\mathrm{g}\) of \(\mathrm{H}_2\mathrm{O}\) were consumed.

Step 1: Balance the given chemical equation

To balance the given chemical equation, we need to adjust the coefficients of the reactants and products so that atoms of each element are equal on both sides of the arrow. The balanced chemical equation for the given reaction is: \[\mathrm{Cl}_{2}\mathrm{O}_{7} + 2\mathrm{H}_{2} \mathrm{O} \rightarrow 2\mathrm{HClO}_{4}\]

Step 2: Calculate the theoretical yield of \(\mathrm{HClO}_{4}\)

Given that actual yield is \(82.0 \% \) which produced \(52.8 \mathrm{~g}\) of \(\mathrm{HClO}_{4}\). We need to find out the theoretical yield. The formula for finding the yield percent is: \[Actual \ Yield \ = \% \ Yield \times Theoretical \ Yield\] Now, let's rearrange the formula to find the theoretical yield. \[Theoretical \ Yield = \frac{Actual \ Yield}{\% \ Yield}\] By plugging the given numbers into the formula, we have: \(Theoretical \ Yield = \frac{52.8 \mathrm{~g}}{0.820}\) \(Theoretical \ Yield = 64.390243902 \mathrm{~g} \approx 64.4 \mathrm{~g}\) So, the theoretical yield of \(\mathrm{HClO}_{4}\) is approximately \(64.4 \mathrm{~g}\).

Step 3: Calculate the consumed grams of \(\mathrm{Cl}_2\mathrm{O}_7\) and \(\mathrm{H}_2\mathrm{O}\)

Now that we have the theoretical yield of \(\mathrm{HClO}_{4}\), we can determine the quantity of consumed \(\mathrm{Cl}_2\mathrm{O}_7\) and \(\mathrm{H}_2\mathrm{O}\) by using the stoichiometry method. First, let's find the mole ratios between \(\mathrm{HClO}_{4}\), \(\mathrm{Cl}_2\mathrm{O}_7\), and \(\mathrm{H}_2\mathrm{O}\) obtained from the balanced chemical equation: \[\frac{2\, moles\,of\, \mathrm{HClO}_{4}}{1\, mole\, of\, \mathrm{Cl}_2\mathrm{O}_7} = \frac{2\, moles\, of\, \mathrm{HClO}_{4}}{2\, moles\, of\, \mathrm{H}_2\mathrm{O}}\] Now, let's find the moles of \(\mathrm{HClO}_{4}\) from the obtained theoretical yield: \[moles\,of\,\mathrm{HClO}_{4} = \frac{64.4\,\mathrm{g}}{100.46\,\frac{\mathrm{g}}{\mathrm{mol}}} = 0.641\,\mathrm{mol}\] Using the mole ratios, we can determine the moles of \(\mathrm{Cl}_2\mathrm{O}_7\) and \(\mathrm{H}_2\mathrm{O}\) consumed in the reaction: \[moles\, of\, \mathrm{Cl}_2\mathrm{O}_7 = 0.641\,\mathrm{mol}\, of\, \mathrm{HClO}_{4} \times \frac{1\, mole\, of\, \mathrm{Cl}_2\mathrm{O}_7}{2\, moles\, of\, \mathrm{HClO}_{4}} = 0.3205\,\mathrm{mol}\] \[moles\, of\, \mathrm{H}_2\mathrm{O} = 0.641\,\mathrm{mol}\, of\, \mathrm{HClO}_{4} \times \frac{2\, moles\, of\, \mathrm{H}_2\mathrm{O}}{2\, moles\, of\, \mathrm{HClO}_{4}} = 0.641\,\mathrm{mol}\] Finally, let's convert moles of \(\mathrm{Cl}_2\mathrm{O}_7\) and \(\mathrm{H}_2\mathrm{O}\) back into grams: \[grams\, of\, \mathrm{Cl}_2\mathrm{O}_7 = 0.3205\,\mathrm{mol} \times 182.903\,\frac{\mathrm{g}}{\mathrm{mol}} = 58.6\,\mathrm{g}\] \[grams\, of\, \mathrm{H}_2\mathrm{O} = 0.641\,\mathrm{mol} \times 18.0153\,\frac{\mathrm{g}}{\mathrm{mol}} = 11.542\,\mathrm{g} \approx 11.5\,\mathrm{g}\] So, about \(58.6\,\mathrm{g}\) of \(\mathrm{Cl}_2\mathrm{O}_7\) and \(11.5\,\mathrm{g}\) of \(\mathrm{H}_2\mathrm{O}\) were consumed in the reaction.

Key Concepts

Theoretical Yield

Theoretical yield is a fundamental concept in stoichiometry. It refers to the maximum amount of product that can be formed from given reactants under ideal conditions. This is calculated based on the balanced chemical equation, assuming that all reactants convert to products with no losses.

In chemical reactions, the actual amount of product obtained (actual yield) is often less than the theoretical yield due to factors like incomplete reactions or side reactions. Understanding the difference between theoretical and actual yield is crucial for efficiency in chemistry.

• To calculate theoretical yield, you need the balanced equation to know the ratios of reactants and products.
• The formula used is \[Theoretical \ Yield = \frac{Actual \ Yield}{\% \ Yield}\].
• In practice, if a reaction is 100% efficient, theoretical and actual yield would be equal.

Knowing how to determine theoretical yield helps in predicting how much product you should expect in perfect conditions. This serves as a benchmark to evaluate the efficiency of a chemical reaction.

Balanced Chemical Equation

Balancing chemical equations is a critical step in stoichiometry. It ensures that the number of atoms for each element is equal on both sides of the reaction. This reflects the Law of Conservation of Mass, which states that mass cannot be created or destroyed in a chemical reaction.

When balancing equations:

• Start with the most complex molecule.
• Adjust coefficients, not subscripts, to balance the atoms.
• Check your work by recounting each type of atom on both sides.

In the given reaction \[\mathrm{Cl}_{2}\mathrm{O}_{7} + 2\mathrm{H}_{2} \mathrm{O} \rightarrow 2\mathrm{HClO}_{4}\], we balance by doubling the \(\mathrm{H}_{2} \mathrm{O}\) reactants to produce the correct amount of \(\mathrm{HClO}_{4}\).
Balancing allows accurate calculations of reactant consumption and product formation using mole ratios.

Mole Ratios

Mole ratios, derived from a balanced chemical equation, are crucial for understanding the relationship between reactants and products. They serve as conversion factors that allow you to switch between moles of different substances involved in the reaction.

From the balanced equation \[\mathrm{Cl}_{2}\mathrm{O}_{7} + 2\mathrm{H}_{2} \mathrm{O} \rightarrow 2\mathrm{HClO}_{4}\], we get several insights:

• 1 mole of \(\mathrm{Cl}_{2}\mathrm{O}_{7}\) yields 2 moles of \(\mathrm{HClO}_{4}\).
• 2 moles of \(\mathrm{H}_{2} \mathrm{O}\) are used to produce 2 moles of \(\mathrm{HClO}_{4}\).
• The mole ratio between \(\mathrm{Cl}_{2}\mathrm{O}_{7}\) and \(\mathrm{H}_{2} \mathrm{O}\) is 1:2.

Understanding mole ratios makes it possible to calculate, step by step, the exact amounts of reactants needed and products formed, ensuring that stoichiometric calculations in chemical reactions are precise.

Chemical Reactions

Chemical reactions involve breaking bonds in reactants and forming new ones in products. The underlying principle is the transformation of matter, which entails energy changes and conservation of atoms.

Key features of a chemical reaction:

• Energy is either absorbed or released, often as heat, light, or electricity.
• Reactions are represented by equations showing the reactants converting into products.
• Types of reactions include synthesis, decomposition, single-replacement, and double-replacement.

In the studied reaction, \(\mathrm{Cl}_{2}\mathrm{O}_{7}\) reacts with \(\mathrm{H}_{2} \mathrm{O}\) to form \(\mathrm{HClO}_{4}\), highlighting a combination-type reaction where two simple molecules combine to form a more complex product.
Grasping how chemical reactions function is essential for predicting outcomes in stoichiometric calculations.