Question

A potential fuel for rockets is a combination of \(\mathrm{B}_{4} \mathrm{H}_{9}\) and \(\mathrm{O}_{2}\) The two react according to the following balanced equation: $$ 2 \mathrm{~B}_{4} \mathrm{H}_{9}(l)+12 \mathrm{O}_{2}(g) \longrightarrow 5 \mathrm{~B}_{2} \mathrm{O}_{3}(s)+9 \mathrm{H}_{2} \mathrm{O}(g) $$ If one tank in a rocket holds \(126 \mathrm{~g} \mathrm{~B}_{4} \mathrm{H}_{9}\) and another tank holds \(192 \mathrm{~g} \mathrm{O}_{2}\), what mass of water can be produced when the entire contents of each tank react together?

Short answer

The mass of water produced when the entire contents of each tank react together is \(81.09\,\text{g}\).

Step-by-step solution

Calculate moles of \(\mathrm{B}_{4}\mathrm{H}_{9}\)

Given mass of \(\mathrm{B}_{4}\mathrm{H}_{9} = 126\,\text{g}\) Molar mass of \(\mathrm{B}_{4}\mathrm{H}_{9} = 4 \times 10.81 + 9 \times 1.01 = 53.24\,\text{g/mol}\) Number of moles of \(\mathrm{B}_{4}\mathrm{H}_{9} = \frac{126\,\text{g}}{53.24\,\text{g/mol}} = 2.366\,\text{mol}\)

Calculate moles of \(\mathrm{O}_{2}\)

Given mass of \(\mathrm{O}_{2} = 192\,\text{g}\) Molar mass of \(\mathrm{O}_{2} = 2 \times 16.00 = 32.00\,\text{g/mol}\) Number of moles of \(\mathrm{O}_{2} = \frac{192\,\text{g}}{32.00\,\text{g/mol}} = 6\,\text{mol}\) Step 2: Determine the limiting reactant

Calculate mole ratios

According to the balanced equation, \(2\,\text{mol}\) of \(\mathrm{B}_{4}\mathrm{H}_{9}\) react with \(12\,\text{mol}\) of \(\mathrm{O}_{2}\) Mole ratio of \(\mathrm{B}_{4}\mathrm{H}_{9} = \frac{2.366\,\text{mol}}{2} = 1.183\) Mole ratio of \(\mathrm{O}_{2} = \frac{6\,\text{mol}}{12} = 0.5\) Since the mole ratio of \(\mathrm{O}_{2}\) is smaller than the mole ratio of \(\mathrm{B}_{4}\mathrm{H}_{9}\), \(\mathrm{O}_{2}\) is the limiting reactant. Step 3: Calculate the mass of water produced

Calculate moles of \(\mathrm{H}_{2}\mathrm{O}\)

According to the balanced equation, \(12\,\text{mol}\) of \(\mathrm{O}_{2}\) produce \(9\,\text{mol}\) of \(\mathrm{H}_{2}\mathrm{O}\) Number of moles of \(\mathrm{H}_{2}\mathrm{O} = \frac{9\,\text{mol}}{12\,\text{mol}} \times 6\,\text{mol} = 4.5\,\text{mol}\)

Calculate mass of \(\mathrm{H}_{2}\mathrm{O}\)

Molar mass of \(\mathrm{H}_{2}\mathrm{O} = 2 \times 1.01 + 1 \times 16.00 = 18.02\,\text{g/mol}\) Mass of \(\mathrm{H}_{2}\mathrm{O} = 4.5\,\text{mol} \times 18.02\,\text{g/mol} = 81.09\,\text{g}\) Hence, the mass of water produced when the entire contents of each tank react together is \(81.09\,\text{g}\).

Key concepts

Understanding Limiting Reactants

When performing chemical reactions, it's crucial to identify the limiting reactant, which is the substance that is completely consumed first and thus determines the amount of products formed. If you imagine cooking a recipe, the ingredient that runs out first limits the final amount of the dish you can make. Similarly, in our rocket fuel reaction, we have to find out whether \textbf{B}\(_4\)\textbf{H}\(_9\) or \textbf{O}\(_2\) is the limiting reactant. This is done by calculating the mole ratios based on the balanced chemical equation and the given masses, and the reactant with the smaller mole ratio is the one that will limit the formation of water and other products.

The identification of the limiting reactant is a fundamental step in stoichiometry problems because it allows us to make precise predictions about the outcome of a chemical reaction. This concept not only guides industrial processes where the efficient use of materials is crucial but also everyday applications such as adjusting recipes in cooking.

The Mole Concept

At the heart of stoichiometry lies the mole concept, which enables chemists to count particles by weighing them. Just as a dozen refers to 12 items, a mole represents 6.022 \times 10^{23} particles (Avogadro's number). To understand the mole concept, think of it as a bridge connecting the microscopic world of atoms and molecules to the macroscopic world we can measure. The molar mass of a substance, which we get from the periodic table, tells us the mass of one mole of that substance.

In the rocket fuel exercise, the mole concept allows us to convert the masses of \textbf{B}\(_4\)\textbf{H}\(_9\) and \textbf{O}\(_2\) to moles so we can use the balanced equation stoichiometrically. This step is essential in calculating the amount of product formed from a given amount of reactant because it standardizes quantities, ensuring they are comparable.

Balancing Chemical Equations

The process of chemical equation balancing is akin to ensuring that both sides of a scale are even. It is based on the Law of Conservation of Mass, which states that mass is neither created nor destroyed in a chemical reaction. Therefore, we must have the same number of each type of atom on both sides of the equation.

In practical terms, for our rocket fuel chemistry problem, balancing the equation is necessary to determine the stoichiometric ratios — that is, the proportional amounts of each reactant and product involved in the reaction. Balanced equations serve as a map for the reaction, providing the mole ratios needed to find the limiting reactant and to calculate the quantity of products such as the mass of water that can be produced.

Rocket Fuel Chemistry

Exploring rocket fuel chemistry is a thrilling application of stoichiometry and illustrates how chemistry propels us into space. Rocket fuels, or propellants, need to react vigorously to produce gases that expand rapidly and thrust the rocket forward by Newton's third law of motion. Each rocket fuel combination has its specific balanced chemical equation, like the one provided for \textbf{B}\(_4\)\textbf{H}\(_9\) and \textbf{O}\(_2\).

These equations allow us to calculate the precise amount of each reactant needed to maximize thrust and determine the byproducts, like water and \textbf{B}\(_2\)\textbf{O}\(_3\), resulting from these high-energy reactions. Understanding these principles of reaction stoichiometry is not only vital for designing efficient rocket engines but also for ensuring the safety and success of space missions.