Problem 12 - Chemistry The Central Science In Si Units

Question

Balance the following equations: (a) \(\mathrm{HClO}_{4}(a q)+\mathrm{P}_{4} \mathrm{O}_{10}(s) \longrightarrow \mathrm{HPO}_{3}(a q)+\mathrm{Cl}_{2} \mathrm{O}_{7}(l)\) (b) \(\mathrm{Au}_{2} \mathrm{~S}_{3}(s)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{Au}(s)+\mathrm{H}_{2} \mathrm{~S}(g)\) (c) \(\mathrm{Ba}_{3} \mathrm{~N}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(a q) \longrightarrow \mathrm{Ba}(\mathrm{OH})_{2}(a q)+\mathrm{NH}_{3}(g)\) (d) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\)

Step-by-Step Solution

Answer

(a) 4:1:4:2, (b) 1:3:2:3, (c) 1:6:3:2, (d) 1:2:2:1:1.

Step 1: Balance equation (a)

The unbalanced equation is:\[ \mathrm{HClO}_{4}(aq)+\mathrm{P}_{4} \mathrm{O}_{10}(s) \longrightarrow \mathrm{HPO}_{3}(aq)+\mathrm{Cl}_{2} \mathrm{O}_{7}(l) \]Identify the number of each type of atom on both sides. Balance the equation starting with the compound that contains the most types of atoms or is involved in a complex molecule. Here, balance the phosphorus (P) and chlorine (Cl) atoms first:- Balance P: There are 4 P atoms in \(\mathrm{P}_{4} \mathrm{O}_{10}\), so we need 4 \(\mathrm{HPO}_{3}\) molecules.- Balance Cl: Since \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) contains 2 Cl atoms, we need 2 molecules of this compound for every 1 of \(\mathrm{P}_{4} \mathrm{O}_{10}\).Finally, balance H and O by adjusting the number of \(\mathrm{HClO}_4\) molecules:\[ 4\mathrm{HClO}_{4}(aq) + \mathrm{P}_{4} \mathrm{O}_{10}(s) \longrightarrow 4\mathrm{HPO}_{3}(aq) + 2\mathrm{Cl}_{2} \mathrm{O}_{7}(l) \]

Step 2: Balance equation (b)

The unbalanced equation is:\[ \mathrm{Au}_{2} \mathrm{~S}_{3}(s) + \mathrm{H}_{2}(g) \longrightarrow \mathrm{Au}(s) + \mathrm{H}_{2} \mathrm{~S}(g) \]Balance the gold (Au) and sulfur (S) atoms:- Balance Au: There are 2 Au atoms in \(\mathrm{Au}_{2} \mathrm{~S}_{3}\), so put a 2 before \(\mathrm{Au}(s)\).- Balance S: There are 3 S atoms in \(\mathrm{Au}_{2} \mathrm{~S}_{3}\), so place a coefficient of 3 in front of \(\mathrm{H}_{2} \mathrm{~S}(g)\).So far, equation looks like:\[ \mathrm{Au}_{2} \mathrm{~S}_{3}(s) + \mathrm{H}_{2}(g) \longrightarrow 2\mathrm{Au}(s) + 3\mathrm{H}_{2} \mathrm{~S}(g) \]Finally, balance hydrogen by placing a coefficient of 3 in front of \(\mathrm{H}_{2}(g)\):\[ \mathrm{Au}_{2} \mathrm{~S}_{3}(s) + 3\mathrm{H}_{2}(g) \longrightarrow 2\mathrm{Au}(s) + 3\mathrm{H}_{2} \mathrm{~S}(g) \]

Step 3: Balance equation (c)

The unbalanced equation is:\[ \mathrm{Ba}_{3} \mathrm{~N}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(aq) \longrightarrow \mathrm{Ba}(\mathrm{OH})_{2}(aq)+\mathrm{NH}_{3}(g) \]Balance the equation starting with barium (Ba) and nitrogen (N):- Balance Ba: There are 3 Ba atoms in \(\mathrm{Ba}_{3} \mathrm{~N}_{2}\), so use 3 \(\mathrm{Ba}(\mathrm{OH})_{2}\).- Balance N: There are 2 N atoms in \(\mathrm{Ba}_{3} \mathrm{~N}_{2}\), so we need 2 \(\mathrm{NH}_{3}\) molecules. Now balance O and H:- Oxygen and Hydrogen: Each \(\mathrm{Ba}(\mathrm{OH})_{2}\) provides 2 O and 2 H for a total of 6 O and 6 H from 3 \(\mathrm{Ba}(\mathrm{OH})_{2}\). Thus, use 6 \(\mathrm{H}_{2} \mathrm{O}\) molecules:\[ \mathrm{Ba}_{3} \mathrm{~N}_{2}(s) + 6\mathrm{H}_{2} \mathrm{O}(aq) \longrightarrow 3\mathrm{Ba}(\mathrm{OH})_{2}(aq) + 2\mathrm{NH}_{3}(g) \]

Step 4: Balance equation (d)

The unbalanced equation is:\[ \mathrm{Na}_{2} \mathrm{CO}_{3}(aq)+\mathrm{HCl}(aq) \longrightarrow \mathrm{NaCl}(aq)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g) \]Balance the sodium (Na) and chlorine (Cl) atoms:- Sodium: There are 2 Na atoms in \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), so place a 2 in front of \(\mathrm{NaCl}\).- Chlorine: As there are 2 Cl atoms needed on the right, place a 2 in front of \(\mathrm{HCl}\). The reaction is now balanced:\[ \mathrm{Na}_{2} \mathrm{CO}_{3}(aq) + 2\mathrm{HCl}(aq) \longrightarrow 2\mathrm{NaCl}(aq) + \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{CO}_{2}(g) \]

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that studies the quantitative relationships in chemical reactions. It revolves around the principle of conservation of mass, which asserts that mass cannot be created or destroyed in a chemical reaction. Therefore, the mass of products in a reaction must equal the mass of reactants. This principle helps chemists understand the proportions of reactants that will produce a given amount of product.

In stoichiometry, we use the term "mole" to represent a specific quantity of particles, usually atoms or molecules. A balanced chemical equation is critical because it provides the mole ratio of reactants and products. This ratio is used to calculate amounts during reactions.

Key steps in stoichiometry involve:

• Converting units of a given substance to moles.
• Using the balanced equation to find the mole ratio.
• Using the mole ratio to calculate moles of the desired substance, and then converting to the required units.

By mastering stoichiometry, students can predict how much of a substance is needed or created in a chemical reaction, making it an essential skill for chemistry studies.

Chemical Reactions

Chemical reactions are the processes by which substances interact to form new substances with different properties. These reactions are the backbone of chemistry and explain how different materials transform and combine.

Chemical reactions can be broadly categorized into synthesis, decomposition, single replacement, and double replacement reactions:

• **Synthesis reactions** involve two or more simple substances combining to form a more complex product.
• **Decomposition reactions** involve a complex molecule breaking down into simpler ones.
• **Single replacement reactions** are processes where one element replaces another in a compound.
• **Double replacement reactions** involve the exchange of ions between two compounds to form new products.

During these reactions, bonds are broken and formed, resulting in energy changes, which can be either exothermic (releasing energy) or endothermic (absorbing energy). Recognizing the type of chemical reaction is crucial, as each type behaves differently and requires specific conditions to occur efficiently.

Equations Balancing Techniques

Balancing chemical equations is an essential skill in chemistry. An unbalanced equation doesn’t reflect accurately what happens in a reaction. The aim is to ensure that the number of each type of atom on the reactant side equals the number of each type of atom on the product side. Here are some techniques to achieve this:

**Identify the Reactants and Products**
Start by clearly identifying the substances involved in the reaction.

**List Elements**
Write down all elements present in the equation, noting the number of atoms of each in the reactants and products.

**Balance One Element at a Time**
This is usually easier done starting with elements that appear in only one reactant and product.

• If an element appears in one molecule, balance it first.
• If a polyatomic ion remains unchanged between the reactants and products, treat it as a single unit.

**Adjust Coefficients**
Modify the coefficients before compounds to balance the atoms—never change the subscripts.

**Check Your Work**
Count the atoms again to ensure every element and the total charge match on both sides.

Building this technique into practice helps students and chemists accurately depict chemical reactions, essential for both academic and practical chemistry applications.