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Given this equation, $$4 \mathrm{~A}_{2}+3 \mathrm{~B} \longrightarrow \mathrm{B}_{3} \mathrm{~A}_{8}$$ use a diagram to illustrate the number of molecules of reactant A and product \(\mathrm{B}_{3} \mathrm{~A}_{8}\) that would be needed/produced from the reaction of six atoms of \(\mathrm{B}\).

Short Answer

Expert verified
8 molecules of \(\mathrm{A}_2\) are needed to produce 2 molecules of \(\mathrm{B}_{3} \mathrm{A}_{8}\) from 6 atoms of \(\mathrm{B}\).

Step by step solution

01

Understand the Stoichiometry of the Reaction

The balanced chemical equation provided is \(4 \mathrm{~A}_{2} + 3 \mathrm{~B} \rightarrow \mathrm{B}_{3} \mathrm{~A}_{8}\). This indicates that 4 molecules of \(\mathrm{A}_2\) react with 3 atoms of \(\mathrm{B}\) to form 1 molecule of \(\mathrm{B}_{3} \mathrm{~A}_{8}\).
02

Determine Reactants for Given B Atoms

We are asked to consider 6 atoms of \(\mathrm{B}\). According to the reaction ratio, 3 atoms of \(\mathrm{B}\) are needed to react with 4 molecules of \(\mathrm{A}_2\). Thus, for 6 atoms of \(\mathrm{B}\), double the number of molecules, \(4 \times 2 = 8\) molecules of \(\mathrm{A}_2\), are needed.
03

Calculate the Product Formed

Since 3 atoms of \(\mathrm{B}\) form 1 molecule of \(\mathrm{B}_3 \mathrm{A}_8\), 6 atoms of \(\mathrm{B}\) will form 2 molecules of \(\mathrm{B}_3 \mathrm{A}_8\) because \(6 / 3 = 2\).
04

Diagram the Reaction

To visualize, draw 8 molecules of \(\mathrm{A}_2\) and 6 atoms of \(\mathrm{B}\). After the reaction, you should depict 2 molecules of \(\mathrm{B}_3 \mathrm{A}_8\) as the product.
05

Review the Diagram

Ensure the diagram accurately reflects the stoichiometry: 8 molecules of \(\mathrm{A}_2\) react with 6 atoms of \(\mathrm{B}\) to produce 2 molecules of \(\mathrm{B}_3 \mathrm{A}_8\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Balanced Chemical Equation
A balanced chemical equation is pivotal in understanding chemical reactions. It shows the exact proportions of reactants and products involved. In our example, the equation is \(4 \mathrm{~A}_{2} + 3 \mathrm{~B} \rightarrow \mathrm{B}_{3} \mathrm{~A}_{8}\). This indicates that 4 molecules of \(\mathrm{A}_{2}\) react with 3 atoms of \(\mathrm{B}\) to form 1 molecule of \(\mathrm{B}_{3} \mathrm{~A}_{8}\). This balance is not just for show! It's a reflection of the Law of Conservation of Mass, which states that matter cannot be created or destroyed.
To ensure accuracy in calculating any chemical reaction, it is crucial to first confirm that the equation is balanced. This means having the same number of each type of atom on both sides of the equation. If you multiply through by a common factor, each term still reflects the same underlying chemistry.
Balancing involves adjusting the coefficients (the numbers in front of the compounds), not the chemical formulas themselves. Remember, the subscripts in a formula are fixed (they are part of the identity of the molecule), whereas the coefficients are variable and adjust to balance the equation.
Reaction Ratios
Understanding reaction ratios is essential in chemical stoichiometry. It enables chemists to determine how much of each reactant is needed and how much of each product will be formed.
In the equation \(4 \mathrm{~A}_{2} + 3 \mathrm{~B} \rightarrow \mathrm{B}_{3} \mathrm{~A}_{8}\), the reaction ratio is evident: 4 molecules of \(\mathrm{A}_{2}\) for every 3 atoms of \(\mathrm{B}\) to produce 1 molecule of \(\mathrm{B}_{3} \mathrm{~A}_{8}\). This ratio provides a direct conversion tool when scaling the reaction up or down.
For instance, if you have 6 atoms of \(\mathrm{B}\), you need to find the equivalent matching quantity of \(\mathrm{A}_{2}\). Since 3 \(\mathrm{B}\) atoms require 4 \(\mathrm{A}_{2}\) molecules, twice the \(\mathrm{B}\) atoms consequently means you double the \(\mathrm{A}_{2}\) molecules as well, resulting in 8 molecules needed. This crucial knowledge allows precise planning of reactions, ensuring efficiency and waste reduction by matching reactant supply closely to demand.
Visualizing Chemical Reactions
Visualizing chemical reactions can greatly aid comprehension, moving from abstract concepts to tangible understanding. Diagramming reactions allows us to see the interaction between molecules and how they transform in chemical processes.
In the given reaction, showing how 8 molecules of \(\mathrm{A}_{2}\) and 6 atoms of \(\mathrm{B}\) react to form 2 molecules of \(\mathrm{B}_{3} \mathrm{~A}_{8}\) can clarify how the actual atom exchange happens in practice.
Drawing diagrams involves arranging the correct number of molecules and representing them before and after the reaction. Through this visual representation, one can easily track atom movement, comprehend complex transformations, and notice any imbalances promptly.
This approach is especially helpful for visual learners. The diagrams help highlight the underlying proportionality in a balanced equation, showing material flow explicitly and reassuring learners with a visual check of their calculations.

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Most popular questions from this chapter

Identify the acid and base used to form these salts, and write the overall neutralization reaction in both complete and net ionic form. (a) \(\mathrm{NaNO}_{2}\) (b) \(\mathrm{CaSO}_{4}\) (c) NaI (d) \(\mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}\)

Potassium hydrogen phthalate, \(\mathrm{KHC}_{8} \mathrm{H}_{4} \mathrm{O}_{4},\) is used to standardize solutions of bases. The acidic anion reacts with bases according to this net ionic equation: $$\mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{2-}(\mathrm{aq})$$ A \(0.902-g\) sample of potassium hydrogen phthalate requires \(26.45 \mathrm{~mL} \mathrm{NaOH}\) to react; determine the molarity of the \(\mathrm{NaOH}\).

Boron forms an extensive series of compounds with hydrogen, all with the general formula \(\mathrm{B}_{x} \mathrm{H}_{y}\). To analyze one of these compounds, you burn it in air and isolate the boron in the form of \(\mathrm{B}_{2} \mathrm{O}_{3}\) and the hydrogen in the form of water. You find that \(0.1482 \mathrm{~g} \mathrm{~B}_{x} \mathrm{H}_{y}\) gives \(0.4221 \mathrm{~g}\) \(\mathrm{B}_{2} \mathrm{O}_{3}\) when burned in excess \(\mathrm{O}_{2} .\) Determine the empirical formula of \(\mathrm{B}_{x} \mathrm{H}_{y}\)

Nitrogen monoxide is oxidized in air to give brown nitrogen dioxide. $$2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})$$ Starting with \(2.2 \mathrm{~mol} \mathrm{NO},\) calculate how many moles and how many grams of \(\mathrm{O}_{2}\) are required for complete reaction. Calculate what mass of \(\mathrm{NO}_{2}\), in grams, is produced.

Classify each of these as an acid or a base. Which are strong and which are weak? What ions are produced when each is dissolved in water? (a) \(\mathrm{KOH}\) (b) \(\mathrm{Mg}(\mathrm{OH})_{2}\) (c) \(\mathrm{HClO}\) (d) \(\mathrm{HBr}\) (e) \(\mathrm{LiOH}\) (f) \(\mathrm{H}_{2} \mathrm{SO}_{3}\)

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