Question

A potential fuel for rockets is a combination of \(\mathrm{B}_{5} \mathrm{H}_{9}\) and \(\mathrm{O}_{2}\) The two react according to the following balanced equation: $$ 2 \mathrm{B}_{5} \mathrm{H}_{9}(l)+12 \mathrm{O}_{2}(g) \longrightarrow 5 \mathrm{B}_{2} \mathrm{O}_{3}(s)+9 \mathrm{H}_{2} \mathrm{O}(g) $$ If one tank in a rocket holds 126 \(\mathrm{g} \mathrm{B}_{5} \mathrm{H}_{9}\) 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

When \(126\,g\) of \(\mathrm{B}_{5}\mathrm{H}_{9}\) and \(192\,g\) of \(\mathrm{O}_{2}\) react, a mass of \(81.072\,g\) of water (\(\mathrm{H}_{2}\mathrm{O}\)) can be produced.

Step-by-step solution

Identify the given information

We have the following information: 1. Balanced chemical equation: $$ 2 \mathrm{B}_{5}\mathrm{H}_{9}(l)+12 \mathrm{O}_{2}(g) \longrightarrow 5 \mathrm{B}_{2}\mathrm{O}_{3}(s)+9 \mathrm{H}_{2}\mathrm{O}(g) $$ 2. Mass of \(\mathrm{B}_{5}\mathrm{H}_{9}\) = \(126\,\text{g}\) 3. Mass of \(\mathrm{O}_{2}\) = \(192\,\text{g}\)

Calculate the moles of each reactant

We need to convert the given masses of \(\mathrm{B}_{5}\mathrm{H}_{9}\) and \(\mathrm{O}_{2}\) into moles using their respective molar masses. Molar mass of \(\mathrm{B}_{5}\mathrm{H}_{9}\): \((5 \times 10.81) + (9 \times 1.008) = 54.05 + 9.072 = 63.122\,\text{g/mol}\) Molar mass of \(\mathrm{O}_{2}\): \(2 \times 16 = 32\,\text{g/mol}\) Moles of \(\mathrm{B}_{5}\mathrm{H}_{9} = \frac{126\,\text{g}}{63.122\,\text{g/mol}} = 1.996\,\text{mol}\) Moles of \(\mathrm{O}_{2} = \frac{192\,\text{g}}{32\,\text{g/mol}} = 6\,\text{mol}\)

Identify the limiting reactant

According to the balanced chemical equation, \(2\,\text{mol}\) of \(\mathrm{B}_{5}\mathrm{H}_{9}\) react with \(12\,\text{mol}\) of \(\mathrm{O}_{2}\). We will now compare the reactants' given moles to their stoichiometric ratio. \(\frac{1.996\,\text{mol}\,\mathrm{B}_{5}\mathrm{H}_{9}}{2} = 0.998\) \(\frac{6\,\text{mol}\,\mathrm{O}_{2}}{12} = 0.5\) The value with the smaller result (0.5) indicates that \(\mathrm{O}_{2}\) is the limiting reactant.

Calculate the mass of water produced

Now that we know \(\mathrm{O}_{2}\) is the limiting reactant, we can determine the mass of water produced using the stoichiometry from the balanced equation: $$ 12\,\text{mol}\, \mathrm{O}_{2} \longrightarrow 9\,\text{mol}\, \mathrm{H}_{2}\mathrm{O} $$ Moles of \(\mathrm{H}_{2}\mathrm{O}\) produced: $$ \frac{9\,\text{mol}\, \mathrm{H}_{2}\mathrm{O}}{12\,\text{mol}\, \mathrm{O}_{2}} \times 6\,\text{mol}\, \mathrm{O}_{2} = 4.5\,\text{mol}\, \mathrm{H}_{2}\mathrm{O} $$ Molar mass of \(\mathrm{H}_{2}\mathrm{O}\): \((2 \times 1.008) + 16 = 18.016\,\text{g/mol}\) Now, we can calculate the mass of water produced: $$ 4.5\,\text{mol}\, \mathrm{H}_{2}\mathrm{O} \times 18.016\,\frac{\text{g}}{\text{mol}} = 81.072\,\text{g} $$

Final Answer

Therefore, a mass of \(81.072\,\text{g}\) of water (\(\mathrm{H}_{2}\mathrm{O}\)) can be produced when the entire contents of each tank react together.

Key concepts

Limiting Reactant

In stoichiometry, the concept of the limiting reactant is crucial in determining the extent of a chemical reaction. The limiting reactant is the reactant that is completely consumed first in a reaction, limiting the amount of product that can be formed. This happens because reactions depend on specific mole ratios that are defined by the balanced chemical equation. To find the limiting reactant, we compare the mole ratio of the actual amounts of reactants used to those required by the balanced chemical equation. For instance, in our rocket fuel example: - According to the balanced equation, 2 moles of \(\mathrm{B}_5\mathrm{H}_9\) require 12 moles of \(\mathrm{O}_2\).- By dividing the actual moles of reactants by the stoichiometric coefficients from the balanced equation (\(\frac{1.996}{2}\) and \(\frac{6}{12}\)), we determine which reactant produces the lesser amount of product when fully consumed.- Thus, \(\mathrm{O}_2\) is identified as the limiting reactant, as it can produce less product compared to the other reactant, \(\mathrm{B}_5\mathrm{H}_9\).

Molar Mass Calculation

Calculating the molar mass of a compound is essential for converting a substance's given mass into moles, which is a crucial step in stoichiometric calculations. Molar mass is the mass of one mole of a given substance, expressed in grams per mole (\(\text{g/mol}\)).For example, consider the molar mass calculation for \(\mathrm{B}_5\mathrm{H}_9\):

• Each boron (B) atom has a molar mass of approximately 10.81 \(\text{g/mol}\), thus 5 boron atoms contribute \(5 \times 10.81 = 54.05 \ \text{g/mol}\)
• Each hydrogen (H) atom has a molar mass of approximately 1.008 \(\text{g/mol}\), making 9 hydrogen atoms contribute \(9 \times 1.008 = 9.072 \ \text{g/mol}\)

Adding these, the molar mass of \(\mathrm{B}_5\mathrm{H}_9\) is \(63.122 \ \text{g/mol}\). Similarly, for \(\mathrm{O}_2\), the calculation is based on two oxygen atoms each with a molar mass of 16 \(\text{g/mol}\), resulting in \(32 \ \text{g/mol}\) for \(\mathrm{O}_2\). These values are then used to convert given masses into moles, setting the stage for further calculations.

Balanced Chemical Equation

The balanced chemical equation is essential for any stoichiometric calculation. It represents the conservation of mass where the number of atoms for each element is equal on both sides of the reaction. This ensures that the reaction follows the principles of the conservation of matter.In our rocket fuel example, the balanced equation is:\[ 2 \ \mathrm{B}_5\mathrm{H}_9(l) + 12 \ \mathrm{O}_2(g) \rightarrow 5 \ \mathrm{B}_2\mathrm{O}_3(s) + 9 \ \mathrm{H}_2\mathrm{O}(g) \]This equation tells us:

• 2 moles of \(\mathrm{B}_5\mathrm{H}_9\) react with 12 moles of \(\mathrm{O}_2\) to produce 9 moles of \(\mathrm{H}_2\mathrm{O}\).
• Both reactants and products have balanced atoms: 10 borons, 18 hydrogens, and 18 oxygens on each side.

Balancing chemical equations ensures that the stoichiometric ratios remain consistent, which is pivotal when determining the reactant quantities and products in a chemical reaction.