Problem 145 - Chemistry An Atoms First Approach

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?

Step-by-Step Solution

Answer

The mass of water that can be produced when the entire contents of each tank react together is 81.09 g.

Step 1: Calculate moles of B5H9 and O2

First, we need to find the moles of B5H9 and O2. Recall that moles = mass/molar mass. The molar mass of B5H9 is (5 × 10.81) + (9 × 1.01) = 54.05 + 9.09 = 63.14 g/mol, and the molar mass of O2 is 2 × 16.00 = 32.00 g/mol. Thus, the moles of B5H9 and O2 are: Moles of B5H9 = (126 g) / (63.14 g/mol) = 2.00 mol Moles of O2 = (192 g) / (32.00 g/mol) = 6.00 mol

Step 2: Determine the limiting reactant

Now, we need to find the limiting reactant, which is the reactant that will be completely consumed first. To do this, we can divide the moles of each reactant by their respective stoichiometric coefficients in the balanced equation: For B5H9: 2.00 mol / 2 = 1.00 mol For O2: 6.00 mol / 12 = 0.50 mol Since 0.50 mol < 1.00 mol, O2 is the limiting reactant.

Step 3: Calculate moles of H2O produced

Using the stoichiometry of the balanced equation, we can now calculate the moles of H2O produced. The balanced equation tells us that 9 mol H2O are produced for every 12 mol O2 that react: Moles of H2O = (9 mol H2O) / (12 mol O2) × 6.00 mol O2 = 4.50 mol H2O

Step 4: Convert moles of H2O to mass

Finally, we can convert the moles of H2O produced into mass using its molar mass. The molar mass of H2O is (2 × 1.01) + 16.00 = 18.02 g/mol: Mass of H2O = (4.50 mol H2O) × (18.02 g/mol) = 81.09 g Thus, a total of 81.09 g of water can be produced when the entire contents of each tank react together.

Key Concepts

Limiting Reactant

Understanding the concept of the limiting reactant is crucial when analyzing chemical reactions, and it basically determines the extent of the reaction. When two or more reactants are combined, the one that is consumed first is the limiting reactant, and it limits the formation of the products. For example, if you're baking cookies and you run out of flour while having plenty of other ingredients left, flour would be your limiting reactant—in the kitchen chemistry of cookie-making!

To identify the limiting reactant in a chemical equation, you need to compare the mole ratio of the reactants used to the ratio from the balanced chemical equation. In the exercise provided, the reaction between \(\mathrm{B}_{5} \mathrm{H}_{9}\) and \(\mathrm{O}_{2}\) was assessed to find that \(\mathrm{O}_{2}\) is the limiting reactant because it runs out first based on stoichiometric coefficients. This means that all further calculations, such as determining the mass of water produced, are based on the amount of \(\mathrm{O}_{2}\) available.

Molar Mass

Molar mass is an essential concept in chemistry that represents the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is a bridge between the mass of a substance and the number of moles since it allows for the conversion between mass in grams and the amount in moles. To calculate molar mass, one simply sums up the atomic masses of all the atoms in a molecule, which are found on the periodic table.

In the context of the problem, we calculated the molar masses of \(\mathrm{B}_{5} \mathrm{H}_{9}\) and \(\mathrm{O}_{2}\) to be 63.14 g/mol and 32.00 g/mol, respectively. These values were then used to convert the given masses of each reactant into moles, thus setting the stage for identifying the limiting reactant and ultimately determining the mass of the product formed.

Stoichiometric Coefficients

Stoichiometric coefficients are the numbers before the molecules in a balanced chemical equation and they tell you the proportion of reactants and products involved in the reaction. These coefficients express the ratio in which the substances react or are produced, thus guiding us on how to convert moles of one substance to moles of another. For instance, a coefficient of 2 before \(\mathrm{B}_{5} \mathrm{H}_{9}\) in the balanced reaction equation signifies that two moles of \(\mathrm{B}_{5} \mathrm{H}_{9}\) will react with twelve moles of \(\mathrm{O}_{2}\) to produce five moles of \(\mathrm{B}_{2} \mathrm{O}_{3}\) and nine moles of water (\(\mathrm{H}_{2} \mathrm{O}\)).

The coefficients are the heart of stoichiometry and allow us to perform accurate calculations in chemical reactions, such as when we determine that 6 moles of \(\mathrm{O}_{2}\) would produce 4.5 moles of \(\mathrm{H}_{2} \mathrm{O}\) by observing the ratios in the balanced equation.

Chemical Reaction Balancing

Balancing chemical reactions is a methodical process to ensure conservation of mass, meaning the number of atoms for each element in the reactants side is equal to that in the products side. When chemical equations are balanced, they reflect the law of conservation of mass, essentially stating that matter cannot be created or destroyed in an ordinary chemical reaction.

In the exercise, the balanced equation \[2 \mathrm{B}_{5} \mathrm{H}_{9}(l)+12 \mathrm{O}_{2}(g) \longrightarrow 5 \mathrm{B}_{2} \mathrm{O}_{3}(s)+9 \mathrm{H}_{2} \mathrm{O}(g)\] indicates the exact ratio in which the reactants combine to form products. Accuracy in balancing equations is paramount as it serves as the foundation for stoichiometric calculations, determining limiting reactants, and predicting the amount of product formed in a chemical reaction. The skill of balancing equations also reinforces a deeper understanding of the reaction mechanisms and stoichiometric relationships within.