Problem 14 - Chemistry: The Central Science

Question

Balance the following equations: (a) \(\mathrm{Ca}_{3} \mathrm{P}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{PH}_{3}(g)\) (b) \(\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (c) \(\mathrm{AgNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \longrightarrow \mathrm{Ag}_{2} \mathrm{CO}_{3}(s)+\mathrm{NaNO}_{3}(a q)\) (d) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)\)

Step-by-Step Solution

Answer

The short version of the answer is: (a) Ca3P2(s) + 6H2O(l) -> 3Ca(OH)2(aq) + 2PH3(g) (b) 2Al(OH)3(s) + 3H2SO4(aq) -> Al2(SO4)3(aq) + 6H2O(l) (c) 2AgNO3(aq) + Na2CO3(aq) -> Ag2CO3(s) + 2NaNO3(aq) (d) 4C2H5NH2(g) + 9O2(g) -> 8CO2(g) + 10H2O(g) + 4N2(g)

Step 1: Equation (a)

To balance the equation, we have to ensure that the number of atoms for each element on the left side matches the number on the right side. Ca3P2(s) + H2O(l) -> Ca(OH)2(aq) + PH3(g) Step 1: Balance Calcium (Ca) atoms 3 Ca atoms on the left, so we need 3 Ca atoms on the right. Ca3P2 + H2O -> 3Ca(OH)2 + PH3 Step 2: Balance Phosphorus (P) atoms 2 P atoms on the left, so we need 2 P atoms on the right. Ca3P2 + H2O -> 3Ca(OH)2 + 2PH3 Step 3: Balance Hydrogen (H) and Oxygen (O) atoms Now we have 6 O atoms on the right side, so we need 6 O atoms on the left side too. Ca3P2 + 6H2O -> 3Ca(OH)2 + 2PH3 The balanced equation for (a) is: Ca3P2(s) + 6H2O(l) -> 3Ca(OH)2(aq) + 2PH3(g)

Step 2: Equation (b)

The equation to balance is: Al(OH)3(s) + H2SO4(aq) -> Al2(SO4)3(aq) + H2O(l) Step 1: Balance Aluminum (Al) atoms There are 2 Al atoms on the right side, so we need 2 Al atoms on the left side. 2Al(OH)3 + H2SO4 -> Al2(SO4)3 + H2O Step 2: Balance Sulfur (S) atoms There are 3 S atoms on the right side, so we need 3 S atoms on the left side. 2Al(OH)3 + 3H2SO4 -> Al2(SO4)3 + H2O Step 3: Balance Hydrogen (H) and Oxygen (O) atoms Now we have 12 H atoms on the left side, so we need 12 H atoms on the right side. 2Al(OH)3 + 3H2SO4 -> Al2(SO4)3 + 6H2O The balanced equation for (b) is: 2Al(OH)3(s) + 3H2SO4(aq) -> Al2(SO4)3(aq) + 6H2O(l)

Step 3: Equation (c)

The equation to balance is: AgNO3(aq) + Na2CO3(aq) -> Ag2CO3(s) + NaNO3(aq) Step 1: Balance Silver (Ag) atoms We have 2 Ag atoms on the right side, so we need 2 Ag atoms on the left side. 2AgNO3 + Na2CO3 -> Ag2CO3 + NaNO3 Step 2: Balance Sodium (Na) atoms We have 2 Na atoms on the left side, so we need 2 Na atoms on the right side too. 2AgNO3 + Na2CO3 -> Ag2CO3 + 2NaNO3 The balanced equation for (c) is: 2AgNO3(aq) + Na2CO3(aq) -> Ag2CO3(s) + 2NaNO3(aq)

Step 4: Equation (d)

The equation to balance is: C2H5NH2(g) + O2(g) -> CO2(g) + H2O(g) + N2(g) Step 1: Balance Carbon (C) atoms We have 2 C atoms on the left side, so we need 2 C atoms on the right side. C2H5NH2 + O2 -> 2CO2 + H2O + N2 Step 2: Balance Hydrogen (H) atoms We have 5 H atoms on the left side, so we need 5 H atoms on the right side. C2H5NH2 + O2 -> 2CO2 + 5/2H2O + N2 Step 3: Balance Nitrogen (N) atoms We have 1 N atom on the left side, so we need 1 N atom on the right side. Note that the equation currently has fractional coefficients; we want integer coefficients, so we will multiply the entire equation by 2. 2C2H5NH2 + 2O2 -> 4CO2 + 5H2O + 2N2 Step 4: Balance Oxygen (O) atoms Now we have 9 O atoms on the right side, so we need 9 O atoms on the left side too. 2C2H5NH2 + 9/2O2 -> 4CO2 + 5H2O + 2N2 Step 5: Get rid of the fractional coefficient Multiply the entire equation by 2 to eliminate the fraction. 4C2H5NH2 + 9O2 -> 8CO2 + 10H2O + 4N2 The balanced equation for (d) is: 4C2H5NH2(g) + 9O2(g) -> 8CO2(g) + 10H2O(g) + 4N2(g)

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry that ensures chemical reactions obey the law of conservation of mass. This means the mass of the reactants must equal the mass of the products. Balancing equations involves making sure the number of atoms for each element is the same on both sides of the equation. To do this, we adjust the coefficients (the numbers in front of compounds), not the subscripts (the numbers within compounds), which are fixed.

• The first step is to write the unbalanced equation.
• Next, balance the atoms for each element one at a time by adjusting coefficients.
• Commonly, it's easiest to balance more complex substances first and leave simpler substances for last, especially diatomic oxygen or hydrogen molecules.
• Finally, double-check all elements are balanced and ensure the coefficients are in the lowest possible ratio.

Consistent practice and a step-by-step approach help in mastering this critical technique. It's like a puzzle, and with practice, you'll become quicker and more adept at it.

Stoichiometry

Stoichiometry involves the quantitative relationships between the amounts of reactants and products in a chemical reaction. It is essentially the math behind chemistry, allowing chemists to predict the quantities of substances consumed and produced in a given reaction.
To perform stoichiometric calculations:

• Begin by writing a balanced chemical equation, as this sets the foundation for all stoichiometric calculations.
• Understand the mole ratio from the balanced equation, which describes the proportions in which substances react and are produced.
• Use molar mass to convert between grams and moles of a substance.
• Solve for the desired quantity - whether it is mass, moles, volume, or molecules.
• Consider limiting reagents, which determine the maximum amount of product possible, as they are completely consumed in the reaction.

Stoichiometry is crucial for chemical engineering, research, and any task that involves converting raw materials into desired products.

Chemistry Education

Chemistry education focuses on teaching the principles, processes, and applications of chemistry in ways that are effectively aligned with educational pedagogy. A substantial part of this education includes learning how to balance chemical equations and understand stoichiometric calculations.
Modern chemistry education encourages hands-on experiments supplemented by theoretical knowledge.

• It aims to build foundational knowledge through interactive and engaging methods, such as laboratories, digital simulations, and real-world problem-solving scenarios.
• Educational resources often include visual aids and comprehensive examples to cater to diverse learning styles.
• Furthermore, conceptual understanding is prioritized over rote memorization, fostering analytical thinking and problem-solving skills.
• Students are encouraged to ask questions and explore chemistry outside the conventional classroom setting.

Chemistry education not only prepares students for future careers in science but also promotes scientific literacy and critical thinking in everyday life.

Chemical Reactions

Chemical reactions are processes in which substances, known as reactants, are transformed into different substances, called products. These reactions involve breaking and forming chemical bonds, which changes the arrangement of atoms.
Types of chemical reactions include:

• Synthesis: Multiple reactants combine to form a single product.
• Decomposition: A single reactant breaks down into multiple products.
• Single Replacement: An element in one compound is replaced by another element.
• Double Replacement: Components of two compounds exchange places, forming two new compounds.
• Combustion: Typically involves a hydrocarbon and oxygen, producing carbon dioxide and water.

During a chemical reaction, energy is either absorbed or released, often in the form of heat, light, or sound. Observing changes in color, temperature, and the formation of precipitates or gas can indicate a chemical reaction. Understanding these processes is essential for applications ranging from cooking to industrial production.