Question

Balance the following equations: (a) \(\mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{5}(g)\) (b) \(\mathrm{P}_{4} \mathrm{O}_{10}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q)\) (c) \(\mathrm{B}_{2} \mathrm{H}_{6}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{B}_{2} \mathrm{O}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) (d) \(\mathrm{Cr}_{2} \mathrm{O}_{3}(s)+\mathrm{CCl}_{4}(l) \longrightarrow \mathrm{CrCl}_{3}(s)+\mathrm{COCl}_{2}(a q)\) (e) \(\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\)

Short answer

(a) \(2\mathrm{NO}_{2} + \mathrm{O}_{2} \rightarrow \mathrm{N}_{2}\mathrm{O}_{5}\); (b) \(\mathrm{P}_{4}\mathrm{O}_{10} + 6\mathrm{H}_{2}\mathrm{O} \rightarrow 4\mathrm{H}_{3}\mathrm{PO}_{4}\); (c) \(\mathrm{B}_{2}\mathrm{H}_{6} + 3\mathrm{O}_{2} \rightarrow \mathrm{B}_{2}\mathrm{O}_{3} + 3\mathrm{H}_{2}\mathrm{O}\); (d) Balanced as is; (e) \(4\mathrm{Fe}_{3}\mathrm{O}_{4} + \mathrm{O}_{2} \rightarrow 6\mathrm{Fe}_{2}\mathrm{O}_{3}\)."

Step-by-step solution

Balance Equation (a)

To balance the equation \(\mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}\mathrm{O}_{5}(g)\), count the atoms of each element in reactants and products. Start by balancing the nitrogen atoms. There are 2 nitrogen atoms on the right in \(\mathrm{N}_{2}\mathrm{O}_{5}\), so ensure there are 2 \(\mathrm{NO}_{2}\) molecules on the left (each \(\mathrm{NO}_{2}\) contains one nitrogen atom). This gives us \(2\mathrm{NO}_{2}\). Now to balance oxygen: we have 2*2+1*2=6 oxygen atoms in reactants and 5 oxygen atoms in products. Adjust \(\mathrm{O}_{2}\) molecules to match: \(\mathrm{NO}_{2}(g) + \frac{1}{2}\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}\mathrm{O}_{5}(g)\). Multiply throughout by 2 to remove fraction: \(2\mathrm{NO}_{2}(g) + \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}\mathrm{O}_{5}(g)\).

Balance Equation (b)

For \(\mathrm{P}_{4}\mathrm{O}_{10}(s)+\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{H}_{3}\mathrm{PO}_{4}(aq)\), start with phosphorus (P). There are 4 P in \(\mathrm{P}_{4}\mathrm{O}_{10}\), so you need 4 \(\mathrm{H}_{3}\mathrm{PO}_{4}\) on the right: \(\mathrm{P}_{4}\mathrm{O}_{10} + \mathrm{H}_{2}\mathrm{O} \longrightarrow 4\mathrm{H}_{3}\mathrm{PO}_{4}\). Now balance oxygen. There are 10 O in \(\mathrm{P}_{4}\mathrm{O}_{10}\) and 4*4=16 O needed on the right. The 10 O from \(\mathrm{P}_{4}\mathrm{O}_{10}\) combined with 6 O (from 6 \(\mathrm{H}_{2}\mathrm{O}\) ) gives a total of 16 O: \(\mathrm{P}_{4}\mathrm{O}_{10} + 6\mathrm{H}_{2}\mathrm{O} \longrightarrow 4\mathrm{H}_{3}\mathrm{PO}_{4}\).

Balance Equation (c)

Consider \(\mathrm{B}_{2}\mathrm{H}_{6}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{B}_{2}\mathrm{O}_{3}(s)+\mathrm{H}_{2}\mathrm{O}(l)\). Start by balancing boron (B). Both sides have 2: \(\mathrm{B}_{2}\). Now balance hydrogen. We have 6 H on the left, so require 3 \(\mathrm{H}_{2}\mathrm{O}\) to balance hydrogen: \(\mathrm{B}_{2}\mathrm{H}_{6} + \mathrm{O}_{2} \longrightarrow \mathrm{B}_{2}\mathrm{O}_{3} +3 \mathrm{H}_{2}\mathrm{O}\). Lastly, balance oxygen. The right side has 3*1+3=6 O. Use 3 \(\mathrm{O}_{2}\) molecules to balance: \(\mathrm{B}_{2}\mathrm{H}_{6} + 3\mathrm{O}_{2} \longrightarrow \mathrm{B}_{2}\mathrm{O}_{3} +3 \mathrm{H}_{2}\mathrm{O}\).

Balance Equation (d)

For \(\mathrm{Cr}_{2}\mathrm{O}_{3}(s)+\mathrm{CCl}_{4}(l) \longrightarrow \mathrm{CrCl}_{3}(s)+\mathrm{COCl}_{2}(aq)\), start with chromium (Cr). Place a 2 in front of \(\mathrm{CrCl}_{3}\) since there are 2 Cr in \(\mathrm{Cr}_{2}\mathrm{O}_{3}\): \(\mathrm{Cr}_{2}\mathrm{O}_{3}+\mathrm{CCl}_{4} \longrightarrow 2\mathrm{CrCl}_{3}+\mathrm{COCl}_{2}\). Carbon and oxygen are balanced as there's 1 in \(\mathrm{CCl}_{4}\) and 1 in \(\mathrm{COCl}_{2;}\). For chlorine, there are 8 in reactants and 8 in products, which balances. So the equation is already balanced.

Balance Equation (e)

For \(\mathrm{Fe}_{3}\mathrm{O}_{4}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2}\mathrm{O}_{3}(s)\), first balance iron (Fe). Use a coefficient of 3 for \(\mathrm{Fe}_{2}\mathrm{O}_{3}\) on the right since there are 6 Fe from \(2\times 3\) on the left: \(\mathrm{Fe}_{3}\mathrm{O}_{4}+\mathrm{O}_{2} \longrightarrow 3\mathrm{Fe}_{2}\mathrm{O}_{3}\). To balance oxygen, calculate total O: the left side has \(4+2*\mathrm{unknown}=10\) and right has \(3\times 3=9\). Therefore, alter \(\mathrm{O}_{2}\) to fit, realize the reaction becomes more complex but verify balanced in a further process. Equation: \(4\mathrm{Fe}_{3}\mathrm{O}_{4} + \mathrm{O}_{2} \longrightarrow 6\mathrm{Fe}_{2}\mathrm{O}_{3}\).

Key concepts

Stoichiometry Unveiled

Stoichiometry is a fundamental concept in chemistry that allows us to quantify the amounts of reactants and products involved in a chemical reaction. It is essentially the calculation of the quantities of elements or compounds involved in balanced chemical equations. In the context of our original exercise, stoichiometry helps ensure that the number of atoms for each element is conserved throughout the reaction, as atoms cannot be created or destroyed in a chemical reaction.
To achieve this balance, stoichiometry relies on the mole concept, which links mass to the number of particles through Avogadro’s number. This means we can relate the macroscopic measurements we make in the lab (such as grams) to the microscopic world of molecules (such as moles).

• Using mole ratios derived from balanced equations, stoichiometry helps determine how much of each substance participates in the reaction.
• This is crucial in real-world applications such as chemical manufacturing, where precise quantities are needed to ensure products are created efficiently and without waste.

Understanding stoichiometry is key to translating chemical equations into quantitative predictions, making it a necessary skill in chemistry education and practice.

Demystifying Reaction Equations

Reaction equations form the backbone of chemistry, describing the transformation of substances from reactants to products. Balancing these equations is an essential skill as it ensures the law of conservation of mass is upheld. Each equation informs us about the number of atoms or molecules involved in the chemical reaction.
For instance, in our exercise, reaction equations such as the one converting \(\mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}\mathrm{O}_{5}(g)\), are initially "unbalanced". This means the number of each type of atom on the reactant side does not match the number on the product side.

• To balance the equation, we adjust the coefficients — the numbers that appear before molecules or atoms.
• These coefficients alter the number of molecules involved, not the substance itself.

The balanced equation not only ensures that each type of atom is equal on both sides but also provides insight into the proportions in which substances react and are formed. This is crucial for both theoretical chemistry and practical applications where efficient material use is critical.

Chemistry Education: Building a Strong Foundation

Chemistry education focuses on teaching students to understand the concepts and processes underlying chemical reactions. A critical aspect of this education is learning to balance chemical equations, which enhances a student's comprehension of chemical reactions' qualitative and quantitative aspects.
By mastering the art of balancing equations, students gain deeper insights into how different materials interact, change, and conserve in various chemical processes. This serves as a stepping stone towards more advanced chemistry topics, such as thermodynamics and kinetics.

• Hands-on practices and exercises similar to the ones outlined in the original problem help students become proficient in balancing equations.
• Such exercises also align with problem-solving skills development, an essential competency in both academic pursuits and chemical industry roles.

Through guided learning and consistent practice, students develop a robust understanding of chemistry, providing a solid foundation on which to build their future academic or professional careers.