Question

For balancing the following reaction: (p) \(\mathrm{CH}_{3} \mathrm{CHO}+\) (q) \(\mathrm{HCHO} \stackrel{\text { (i) NaOH }}{\text { (ii) } \mathrm{H}^{\prime}}{\longrightarrow}\) (r) \(\mathrm{C}\left(\mathrm{CH}_{2} \mathrm{OH}\right)_{4}+(\mathrm{s}) \mathrm{HCOOH}\) \(\mathrm{p}, \mathrm{q}, \mathrm{r}, \mathrm{s}\) are respectively: (A) \(2,2,1,1\) (B) \(3,2,1,3\) (C) \(1,4,1,1\) (D) \(1,4,1,0\)

Short answer

The short answer based on the given step-by-step solution is: The correct coefficients for the balanced chemical equation are (C) \(1, 4, 1, 0\).

Step-by-step solution

Write down the given unbalanced equation

Start by writing down the given unbalanced chemical equation: \(p \mathrm{CH}_{3}\mathrm{CHO} + q \mathrm{HCHO} \rightarrow r \mathrm{C}\left(\mathrm{CH}_{2}\mathrm{OH}\right)_{4} + s\mathrm{HCOOH}\)

Balancing the carbons

First, we balance the number of carbon atoms on both sides of the equation. On the right side, we have "r" (one C atom from C(CH₂OH)₄) and "s" (one C atom from HCOOH). On the left side, we have "p" (two C atoms from CH₃CHO) and "q" (one C atom from HCHO). Balancing the carbons gives us: \(2p + q = r + s\)

Balancing the hydrogens

Next, we balance the number of hydrogen atoms on both sides of the equation. On the right side, we have "4r" (eight H atoms from C(CH₂OH)₄) and "2s" (two H atoms from HCOOH). On the left side, we have "4p" (four H atoms from CH₃CHO) and "2q" (two H atoms from HCHO). Balancing the hydrogens gives us: \(4p + 2q = 8r + 2s\)

Balancing the oxygens

Finally, we balance the number of oxygen atoms on both sides of the equation. On the right side, we have "4r" (four O atoms from C(CH₂OH)₄) and "2s" (two O atoms from HCOOH). On the left side, we have "p" (one O atom from CH₃CHO) and "q" (one O atom from HCHO). Balancing the oxygens gives us: \(p + q = 4r + 2s\)

Solve the system of equations

Now we have a system of three equations: 1. \(2p + q = r + s\) 2. \(4p + 2q = 8r + 2s\) 3. \(p + q = 4r + 2s\) From equation (1), we get: \(r + s = 2p + q = 2(p + q)\) Using equation (3) to substitute for \(p + q\): \(r + s = 2(4r + 2s)\) Simplifying: \(r + s - 8r - 4s = 0\) \(-7r - 3s = 0\) From equation (2), we also get: \(4p + 2q = 8r + 2s\) \(2p + q = 4r + s\) Using equation (1) to substitute for \(2p + q\): \(r + s = 4r + s\) \(r = 3r\) This implies that \(r = 1\). Now we have the new equation: \(-7(1) - 3s = 0\) \(-7 = 3s\) \(s = -\dfrac{7}{3}\) Since stoichiometric coefficients must be whole numbers, the obtained value for 's' is not possible. We go back to our equations (1) and (3): \(2p + q = r + s\) \(p + q = 4r + 2s\) Since the coefficients should be whole numbers, let's consider the simplest solution when \(r = 1\) and \(s = 0\), which would result in: \(2p + q = 1\) \(p + q = 4\) Solving the above two equations, we get: \(p = 1\) \(q = 4\) Thus, the final balanced chemical equation is: \(1 \mathrm{CH}_{3}\mathrm{CHO} + 4 \mathrm{HCHO} \rightarrow 1 \mathrm{C}\left(\mathrm{CH}_{2}\mathrm{OH}\right)_{4}\)

Compare and find the answer

Now that we've found the balanced equation, we can compare it with our choices and identify the correct answer. The final coefficients match option (C): \(p = 1\), \(q = 4\), \(r = 1\), and \(s = 0\). The correct answer is: (C) \(1, 4, 1, 0\)

Key concepts

Chemical Equations

Chemical equations represent chemical reactions using symbols and formulas for the substances involved. The left-hand side of the equation shows the reactants, while the right-hand side lists the products. Balancing these equations is crucial for demonstrating the law of conservation of mass, which states that mass cannot be created or destroyed in a chemical reaction. This means the number of atoms for each element must be the same on both sides of the equation.

To balance an equation, adjust the coefficients (the numbers in front of molecules or atoms) without changing the chemical nature of the substances involved. For example, balancing makes sure the number of carbon, hydrogen, and oxygen atoms are equal on both sides of the reaction. A properly balanced chemical equation not only shows the relative quantities of reactants and products but also sets the stage for stoichiometric calculations.

Understanding chemical equations lays the foundation for predicting the outcomes of reactions and quantifying the amount of reactants needed or products formed.

Stoichiometry

Stoichiometry is a section of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It uses the balanced chemical equation to calculate the amounts of various substances involved in a reaction. This concept hinges on the law of conservation of mass and the principle that matter is neither created nor destroyed during chemical reactions.

A stoichiometric calculation typically begins with a balanced chemical equation, from which stoichiometric coefficients are drawn. These coefficients directly relate the moles of each substance involved. For example, if the equation says you need 1 mole of A for every 2 moles of B to produce 1 mole of C, these numbers are key for calculating how much of each substance will react or be produced.

• Identify the known and unknown quantities in terms of moles with the help of stoichiometric coefficients from the balanced equation.
• Convert known mass or volume amounts into moles, if necessary, using molar masses or gas laws.
• Utilize the ratio provided by the balanced equation to determine the moles or mass of the unknown.

Stoichiometry is fundamental for designing reactions and processes, ensuring materials are used efficiently and waste is minimized.

Organic Chemistry Reactions

Organic chemistry focuses on the study of carbon-containing compounds, which are the basis of life. Reactions in this field often involve the making and breaking of covalent bonds between atoms. Distinct from inorganic reactions, organic reactions can involve intricate pathways and mechanisms.

One type of organic reaction is the synthesis of complex molecules from simpler ones, often involving condensation reactions where water is released. An example is the reaction provided: a series of transformations involving acetaldehyde (\(\mathrm{CH}_3\mathrm{CHO}\)) and formaldehyde (\(\mathrm{HCHO}\)) to form a larger compound.

Reactivity in organic chemistry is influenced by various factors, including the structure of molecules, presence of functional groups (specific groupings of atoms that impart certain reactivity), and the surrounding environment. Understanding these aspects helps predict the outcome of reactions and design new pathways for creating novel compounds.

• Functional groups such as alcohols, carboxyls, and amines play crucial roles in determining the nature of organic reactions.
• Reaction types include addition, substitution, and elimination, each involving different structural changes.
• Organic reactions are often catalyzed by acids, bases, or enzymes, which increase the reaction rate by lowering the activation energy.

Mastery of organic chemistry reactions opens doors to innovations in pharmaceuticals, materials science, and biochemistry.