Question

Give a step-by-step procedure for estimating the number of grams of \(A\) required to completely react with 10 moles of \(B\), given the following information: \(A\) and \(B\) react to form \(A_{5} B_{2}\) \(A\) has a molecular mass of \(100.0 . B\) has a molecular mass of 200.0 There are \(6.02 \times 10^{23}\) molecules of \(A\) in a mole of \(A\). Which of these pieces of information weren't needed?

Short answer

2500 grams of A are required to react with 10 moles of B. The information about the number of molecules in a mole (Avogadro's number) wasn't needed.

Step-by-step solution

Write the balanced chemical equation

From the given information, the chemical equation for the reaction between A and B forming \(A_{5}B_{2}\) can be written as: \[\begin{equation} 5A + 2B \rightarrow A_{5}B_{2} \end{equation}\] This equation tells us that 5 moles of A react with 2 moles of B to produce one mole of \(A_{5}B_{2}\).

Calculate the moles of A needed

To completely react with 10 moles of B, we need to determine the moles of A required using the stoichiometry of the reaction. Since the ratio of A to B is 5:2, we perform cross multiplication to find the moles of A needed: \[\begin{equation} \frac{5 \, \text{moles of A}}{2 \, \text{moles of B}} = \frac{x \, \text{moles of A}}{10 \, \text{moles of B}} \end{equation}\] Solving for x, we get: \[\begin{equation} x = \frac{5}{2} \times 10 = 25 \, \text{moles of A} \end{equation}\] This means that 25 moles of A are required to react completely with 10 moles of B.

Calculate the mass of A required

To find the number of grams of A required, we use the molecular mass of A. As one mole of A has a mass of 100.0 grams, the mass of 25 moles of A can be calculated by: \[\begin{equation} \text{Mass of A} = 25 \, \text{moles of A} \times 100.0 \, \text{grams/mole} = 2500 \, \text{grams} \end{equation}\] Therefore, 2500 grams of A are required to react completely with 10 moles of B.

Excess Information Identification

The piece of information about the number of molecules in a mole of A (Avogadro's number) wasn't necessary for this calculation because we were dealing with moles and masses, not individual molecules.

Key concepts

Chemical Reaction Stoichiometry

Chemical reaction stoichiometry is crucial for scientists and engineers to predict the quantities of reactants needed and products formed in a chemical reaction. It is essentially the bookkeeping of chemistry, ensuring that atoms are conserved in any given reaction according to the Law of Conservation of Mass.

Using stoichiometry, we can translate the qualitative information provided by a balanced chemical equation into quantitative data. For instance, in our exercise, the reaction between substances A and B is outlined by a stoichiometric equation, indicating the exact ratio of reactants needed for the reaction to proceed without any leftover reactants. In order to calculate the amount of A needed to completely react with 10 moles of B, we first make sure our reaction equation is balanced, then use stoichiometric coefficients to find out the proportion of moles of A to moles of B.

For educational purposes, it's crucial to understand that these coefficients reflect molar ratios and can be used to convert between the moles of one substance and the moles of another. Thus, stoichiometry serves not only in calculating the amounts of reactants and products but also aids in critical thinking when considering the efficiency and yield of chemical processes.

Molar Mass Calculation

In the realm of chemistry, the molar mass is a pivotal concept that connects the macroscopic world we can measure to the microscopic world of atoms and molecules. It's defined as the mass of one mole of a substance, usually expressed in grams per mole (g/mol).

The molar mass gives us a way to convert between mass and moles, thus anchoring stoichiometry to tangible quantities. In our example, where we calculated the amount of reactant A required to fully engage with 10 moles of reactant B, we have used the molar mass of A, given as 100.0 g/mol, to determine the total mass needed. By multiplying the number of moles (25 moles, as calculated from our equation) by the molar mass, the resulting product gives us the required grams of A.

Understanding how to calculate molar mass is not only fundamental for this example but is an essential skill for any student tackling chemistry problems. By being precise in these calculations, students gain an appreciation of how molecular properties relate to the macroscopic quantities they can experiment with in the laboratory setting.

Balanced Chemical Equation

The balanced chemical equation serves as the foundation for all stoichiometric calculations. It's a representation of a chemical reaction where the number of atoms for each element is the same on both the reactant and product sides, obeying the Law of Conservation of Mass. Balancing the equation is a necessary first step before any reaction-related calculation can take place.

In our exercise, the balanced equation was derived from the given reaction: 5A + 2B → A₅B₂. This shows a direct relationship between reactants A and B, and their stoichiometric coefficients (5 and 2, respectively) tell us how many moles of each reactant are needed to form the product without any excess. Balancing equations is not only a matter of making the numbers match but understanding the mole ratio, which then informs us how much of each reactant is consumed and how much product is formed during the reaction.

Students must learn how to balance chemical equations as it underpins their ability to predict the outcomes of reactions accurately. Whether it's a simple synthesis or a more complex reaction, mastering this skill ensures students can perform stoichiometric conversions confidently and efficiently.