Problem 78 - Chemistry: Principles and Reactions

Question

The Space Shuttle uses aluminum metal and ammonium perchlorate in its reusable booster rockets. The products of the reaction are aluminum oxide, aluminum chloride, nitrogen oxide gas, and steam. The reaction mixture contains \(7.00 \mathrm{~g}\) of aluminum and \(9.32 \mathrm{~g}\) of ammonium perchlorate. (a) Write a balanced equation for the reaction. (b) What is the theoretical yield of aluminum oxide? (c) If \(1.56 \mathrm{~g}\) of aluminum oxide is formed, what is the percent yield? (d) How many grams of excess reactant are unused? (Assume \(100 \%\) yield.)

Step-by-Step Solution

Answer

Answer: The percent yield of aluminum oxide is 15.4%, and the mass of excess aluminum unused is 3.44 g.

Step 1: Write a balanced chemical equation

The balanced chemical equation for the reaction is: \(\mathrm{10 Al_{(s)} + 6 NH_4ClO_4_{(s)}\rightarrow 7.5Al_2O_3_{(s)} + 3AlCl_3_{(s)} + 6NO_{(g)} + 12H_2O_{(g)}}\)

Step 2: Calculate the moles of each reactant

Using the molar mass of each reactant, we can calculate the moles: For aluminum: \(\mathrm{Moles\ Al} = \frac{7.00\ g}{26.98\ g/mol} = 0.259\ mol\) For ammonium perchlorate: \(\mathrm{Moles\ NH_4ClO_4} = \frac{9.32\ g}{117.49\ g/mol} = 0.0792\ mol\)

Step 3: Determine the limiting reactant

To find the limiting reactant, we need to compare the mole ratios to the balanced equation coefficients: Mole ratio Al: \(\frac{0.259\ mol}{10} = 0.0259\) Mole ratio \(\mathrm{NH_4ClO_4}\): \(\frac{0.0792\ mol}{6} = 0.0132\) Since the mole ratio of ammonium perchlorate is smaller, it is the limiting reactant.

Step 4: Calculate the theoretical yield of aluminum oxide

Using the limiting reactant and stoichiometry, we can calculate the theoretical yield of aluminum oxide: \(\mathrm{Moles\ Al_2O_3} = \frac{7.5 \ mol\ Al_2O_3}{6 \ mol\ NH_4ClO_4} \times 0.0792 \ mol\ NH_4ClO_4 = 0.0990\ mol\ Al_2O_3\) Now, we'll convert the moles of aluminum oxide to grams: \(\mathrm{Theoretical\ yield\ Al_2O_3} = 0.0990\ mol \times 101.96\ g/mol = 10.1\ g\)

Step 5: Determine the percent yield

Using the actual mass of aluminum oxide formed and the theoretical yield, we can calculate the percent yield: \(\mathrm{Percent\ yield} = \frac{1.56\ g}{10.1\ g} \times 100\% = 15.4\%\)

Step 6: Calculate the mass of the excess reactant unused

Now we can determine the mass of aluminum used, assuming 100% yield: \(\mathrm{Moles\ Al\ used} = \frac{10\ mol\ Al}{6\ mol\ NH_4ClO_4} \times 0.0792\ mol\ NH_4ClO_4 = 0.132\ mol\ Al\) \(\mathrm{Mass\ Al\ used} = 0.132\ mol \times 26.98\ g/mol = 3.56\ g\) Since we started with \(7.00\ g\) of aluminum, the mass of excess aluminum unused is: \(\mathrm{Excess\ Al} = 7.00\ g - 3.56\ g = 3.44\ g\) In conclusion, the answers are: a) The balanced equation is: \(\mathrm{10 Al_{(s)} + 6 NH_4ClO_4_{(s)}\rightarrow 7.5Al_2O_3_{(s)} + 3AlCl_3_{(s)} + 6NO_{(g)} + 12H_2O_{(g)}}\) b) The theoretical yield of aluminum oxide is \(10.1\ g\). c) The percent yield is \(15.4\%\). d) The mass of excess aluminum unused is \(3.44\ g\).

Key Concepts

Chemical Reactions

Understanding chemical reactions is essential for anyone studying chemistry. These reactions involve the transformation of substances into different products through the breaking and forming of chemical bonds.

For instance, in the Space Shuttle example, we examined a complex reaction where aluminum metal reacts with ammonium perchlorate to produce aluminum oxide, aluminum chloride, nitrogen oxide, and steam. Writing a balanced chemical equation is crucial as it ensures that the number of atoms for each element is the same on both sides of the equation, adhering to the law of conservation of mass. This equation provides the necessary ratios of reactants and products to perform stoichiometry calculations.

By understanding chemical reactions, students can predict the formation of products, calculate the amounts needed or produced, and comprehend the fundamental processes that drive physical phenomena.

Theoretical Yield

In stoichiometry, the theoretical yield is the maximum amount of product that could be generated from a given amount of reactant, according to the stoichiometric coefficients in the balanced equation.

To illustrate, when considering the Space Shuttle's booster rockets, after identifying the limiting reactant as ammonium perchlorate, the theoretical yield of aluminum oxide is calculated. This calculation is done by first determining how many moles of reactant will react and then converting these moles to grams using the molecular weight of the product. The theoretical yield informs us of the upper limit of product that can be expected in a perfect scenario, where all reactant is converted with no losses.

Percent Yield

In a real-world laboratory setting, reactions rarely proceed perfectly, and practical limitations result in a lesser amount of product than the theoretical yield. The percent yield is a measurement of the efficiency of a chemical reaction, expressing the actual yield as a percentage of the theoretical yield.

For example, from the Space Shuttle booster rocket chemical reaction, if only 1.56 grams of aluminum oxide is collected when the theoretical yield is 10.1 grams, the percent yield is calculated to be 15.4%. Understanding the percent yield helps in evaluating the effectiveness of a chemical process and identifying possible improvements for industrial or experimental methodologies.

Limiting Reactant

In every chemical reaction involving two or more reactants, one reactant will be used up first, preventing further product from forming. This is known as the limiting reactant. Identifying the limiting reactant is crucial as it determines the theoretical yield of the reaction.

In our Space Shuttle example, ammonium perchlorate is determined to be the limiting reactant by comparing the mole ratios provided by the reaction's stoichiometry. Knowing which reactant limits the reaction allows for precise calculation of the expected amount of products and informs decisions about cost-effectiveness in situations where reactants vary in price or availability. It also helps researchers in determining the excess of other reactants, which can be recovered or recycled.