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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) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\)

Short Answer

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

Step by step solution

01

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:\[ 4\mathrm{HClO}_{4}(aq) + \mathrm{P}_{4} \mathrm{O}_{10}(s) \longrightarrow 4\mathrm{HPO}_{3}(aq) + 2\mathrm{Cl}_{2} \mathrm{O}_{7}(l) \]
02

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 \(\mathrm{H}_{2}(g)\):\[ \mathrm{Au}_{2} \mathrm{~S}_{3}(s) + 3\mathrm{H}_{2}(g) \longrightarrow 2\mathrm{Au}(s) + 3\mathrm{H}_{2} \mathrm{~S}(g) \]
03

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:\[ \mathrm{Ba}_{3} \mathrm{~N}_{2}(s) + 6\mathrm{H}_{2} \mathrm{O}(aq) \longrightarrow 3\mathrm{Ba}(\mathrm{OH})_{2}(aq) + 2\mathrm{NH}_{3}(g) \]
04

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:\[ \mathrm{Na}_{2} \mathrm{CO}_{3}(aq) + 2\mathrm{HCl}(aq) \longrightarrow 2\mathrm{NaCl}(aq) + \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{CO}_{2}(g) \]

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Key Concepts

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

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.

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Most popular questions from this chapter

An element \(X\) forms an iodide \(\left(\mathrm{XI}_{3}\right)\) and a chloride \(\left(\mathrm{XCl}_{3}\right)\) The iodide is quantitatively converted to the chloride when it is heated in a stream of chlorine: $$ 2 \mathrm{XI}_{3}+3 \mathrm{Cl}_{2} \longrightarrow 2 \mathrm{XCl}_{3}+3 \mathrm{I}_{2} $$ If \(0.5000 \mathrm{~g}\) of \(\mathrm{XI}_{3}\) is treated with chlorine, \(0.2360 \mathrm{~g}\) of \(\mathrm{XCl}_{3}\) is obtained. (a) Calculate the atomic weight of the element \(\mathrm{X}\). (b) Identify the element X.

The reaction between potassium superoxide, \(\mathrm{KO}_{2}\), and \(\mathrm{CO}_{2}\), $$ 4 \mathrm{KO}_{2}+2 \mathrm{CO}_{2} \longrightarrow 2 \mathrm{~K}_{2} \mathrm{CO}_{3}+3 \mathrm{O}_{2} $$ is used as a source of \(\mathrm{O}_{2}\) and absorber of \(\mathrm{CO}_{2}\) in selfcontained breathing equipment used by rescue workers. (a) How many moles of \(\mathrm{O}_{2}\) are produced when \(0.400 \mathrm{~mol}\) of \(\mathrm{KO}_{2}\) reacts in this fashion? (b) How many grams of \(\mathrm{KO}_{2}\) are needed to form \(7.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\) ?

Propenoic acid, \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{2},\) is a reactive organic liquid that is used in the manufacturing of plastics, coatings, and adhesives. An unlabeled container is thought to contain this liquid. A 0.275 -g sample of the liquid is combusted to produce \(0.102 \mathrm{~g}\) of water and \(0.374 \mathrm{~g}\) carbon dioxide. Is the unknown liquid propenoic acid? Support your reasoning with calculations.

The fat stored in a camel's hump is a source of both energy and water. Calculate the mass of \(\mathrm{H}_{2} \mathrm{O}\) produced by the metabolism of \(1.0 \mathrm{~kg}\) of fat, assuming the fat consists entirely of tristearin \(\left(\mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_{6}\right)\), a typical animal fat, and assuming that during metabolism, tristearin reacts with \(\mathrm{O}_{2}\) to form only \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\).

(a) You are given a cube of silver metal that measures 1.000 \(\mathrm{cm}\) on each edge. The density of silver is \(10.5 \mathrm{~g} / \mathrm{cm}^{3}\). How many atoms are in this cube? (b) Because atoms are spherical, they cannot occupy all of the space of the cube. The silver atoms pack in the solid in such a way that \(74 \%\) of the volume of the solid is actually filled with the silver atoms. Calculate the volume of a single silver atom. (c) Using the volume of a silver atom and the formula for the volume of a sphere, calculate the radius in angstroms of a silver atom.

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