07 Chemical Equations Concept Overview
CHEMICAL EQUATIONS | CONCEPT OVERVIEW The TOPIC of CHEMICAL EQUATIONS is not directly provided to us in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. However, it is of upmost importance that we understand the fundamental concepts and applications revolving around this subject independent of the NCEES Supplied Reference Handbook.
CONCEPT INTRO: A CHEMICAL REACTION is a PROCESS in which a SUBSTANCE or SUBSTANCES is CHANGED into one or more NEW SUBSTANCES. A CHEMICAL EQUATION represents the CHANGE that occurs during a CHEMICAL REACTION. A CHEMICAL EQUATION is a written REPRESENTATION of a CHEMICAL REACTION, using the SYMBOLS of each ELEMENT to indicate the AMOUNT of SUBSTANCE of each REACTANT and PRODUCT. An EXAMPLE of a simple CHEMICAL REACTION is the REACTION in which hydrogen gas (π»2 ) and oxygen gas (π2 ) COMBINE to product liquid water (π»2 0).
The GENERAL FORMULA for a CHEMICAL EQUATION is not directly provided to us in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Made with by Prepineer | Prepineer.com
However, it is of upmost importance that we understand the fundamental concepts and applications revolving around this subject independent of the NCEES Supplied Reference Handbook. The GENERAL FORM of a CHEMICAL REACTION demonstrates the component of a CHEMICAL PROCUESS by defining the REACTS that REACT and YIELD to produce PRODUCTS. π ππππ‘πππ‘π β πππππ’ππ‘π The YIELDS symbol " β " separates the REACTS from the PRODUCTS in a CHEMICAL EQUATION. A ONE-WAY YIELD symbol ββ " suggests that the REACTION occurs in only one direction. A TWO-WAY YIELDS symbol βββ indicates that the REACTION can occur REVERSIBLY, in both DIRECTIONS. The LEFT SIDE of the CHEMICAL EQUATION shows the REACTANTS, which are the ELEMENTS or COMPOUNDS that REACT in order for a CHEMICAL REACTION to OCCUR. The RIGHT SIDE of the CHEMICAL EQUATIONS shows the PRODUCTS, which are the ELEMENTS or COMPOUNDS that are PRODUCED as a RESULT of the CHEMICAL REACTION of the REACTANTS.
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In the earlier example, of hydrogen (π»2 ) gas and oxygen (π2 ) gas combining to form liquid water, HYDROGEN and OXYGEN would be the REACTANTS, and a WATER molecule would be the PRODUCT. 2π»2(π) + π2(π) β π»2 π(π) The STATE of MATTER of each SUBSTANCE is represented by an ABBREVIATION in the SUBSCRIPT of each REACTANT and PRODUCT. The NOTATE for each STATE of MATTER is as follows: β’ πΊππ = π(π) β’ πΏπππ’ππ = π(π) β’ πππππ = π(π ) β’ π΄ππ’πππ’π ππππ’π‘πππ = π(ππ) An AQUEOUS SOLUTION is a SOLID that is DISSOLVED in a LIQUID SOLUTION, resulting in a DILUTED SOLUTION. For example, given a sample of the oxygen that we breathe in as a gas, we would EXPRESS oxygen as: π2 (π) In a BALANCED CHEMICAL EQUATION, the NUMBER of ATOMS of each ELEMENT is EQUAL on both sides of the CHEMICAL EQUATION. The TOPIC of LAW OF CONSERVATION OF MASS is not directly provided to us in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Made with
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However, it is of upmost importance that we understand the fundamental concepts and applications revolving around this subject independent of the NCEES Supplied Reference Handbook. It is important to BALANCE CHEMICAL EQUATIONS in order to follow the LAW OF CONSERVATION MASS, which states that all the MASS or ATOMS present at the BEGINNING of a REACTION must be PRESENT in the FORM of a PRODUCT. Simply speaking, the LAW OF CONSERVATION states that there must be an EQUAL number of ATOMS of each ELEMENT in the REACTANTS as in the PRODUCTS of the CHEMICAL EQUATION. Before we BALANCE a CHEMICAL EQUATION, letβs look at the GENERAL FORM of the SKELETON EQUATION for any CHEMICAL EQUATION:
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β’ A, B, C, and D are the GROUPINGS, which represent the NUMBER of ATOMS in the REACTANT or PRODUCTS. A GROUPING is typically an ELEMENT, MOLECULE, or COMPOUND. β’ N, M, O, and P are the COEFFICIENTS of each GROUPING, which REPRESENT the NUMBER of MOLS for each ELEMENT, COMPOUND, or MOLECULE. β’ W, X, Y, and Z are the SUBSCRIPTS, which are part of the CHEMICAL FORMULA for a particular GROUPING and cannot be changed. When approaching a CHEMICAL EQUATION, it is important you understand the difference between COEFFICIENTS and SUBSCRIPTS. The COEFFICIENT is a WHOLE NUMBER INTEGER placed in front of an ELEMENT, COMPOUND, or MOLECULE, and denotes the AMOUNT of MOLES for a particular ELEMENT, COMPOUND, or MOLECULE. The SUBSCRIPT is placed behind the ELEMENT, MOLECULE, or COMPOUND and denotes the NUMBER of ATOMS in that GROUPING. To BALANCE an EQUATION, COEFFICIENTS are used to MODIFY the NUMBER of MOLES of REACTANTS and/or PRODUCT, such that MASS is CONSERVED on BOTH SIDE of the CHEMICAL EQUATION. CHANGING the COEFFICIENT changes the TOTAL NUMBER of MOLES of that ELEMENT or MOLECULE. The SUBSCRIPT, however cannot be changed, as altering a SUBSCRIPT would change the MOLECULE itself.
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The TOPIC of DIATOMIC MOLECULES is not directly provided to us in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. However, it is of upmost importance that we understand the fundamental concepts and applications revolving around this subject independent of the NCEES Supplied Reference Handbook. A DIATOMIC MOLECULE is a MOLECULE made of only TWO ATOMS, of either the SAME or different CHEMICAL ELEMENTS. Some ELEMENTS can only exist as DIATOMIC MOLECULES and are not physically able to exist as a single ATOM, but as a PAIR of ATOMS that share ELECTRONS to reach the nearest OCTET. The SEVEN ELEMENTS that exist only as DIATOMIC MOLECULES are hydrogen, nitrogen, oxygen, fluorine, chlorine, iodine, and bromine. You can easily remember the DIATOMIC ELEMENTS with the phrase:
πave πo π ear πf πce πold πeer Where: β’ H represents Hydrogen (π»2 ) β’ N represents Nitrogen (π2 ) β’ F represents Fluorine (πΉ2 ) β’ O represents Oxygen (π2 ) Made with
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β’ I represents Iodine (πΌ2 ) β’ C represents Chlorine (πΆπ2 ) β’ B represents Bromine (π΅π2 )
As with any EQUATION in engineering, we can BALANCE CHEMICAL EQUATIONS using a simple 4 STEP PROCESS. 1. Write Out The Problem Statement: IDENTIFY all REACTANTS and PRODUCTS and WRITE their correct FORMULAS on each SIDE of the CHEMICAL EQUATION. It is crucial to correctly TRANSLATE the PROBLEM STATEMENT to correctly WRITE out the CHEMICAL EQUATION. For example, a sample of sodium is mixed with chlorine gas to form sodium chloride. We can write the SKELETON EQUATION for this CHEMICAL REACTION as:
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ππ + πΆπ2 β πππΆπΏ
2. Balance The Atoms For Each Element: Begin balancing the equation by trying different COEFFICIENTS to ADJUST the NUMBER of ATOMS of each ELEMENT to be the SAME on both sides of the CHEMICAL EQUATION. We can change the COEFFICIENTS (the numbers preceding the formulas), but not the SUBSCRIPTS (the number within chemical formulas), as changing the subscripts would change the IDENTITY of the SUBSTANCE. 1 1 ππ + πΆπ2 β 1 ππ πΆπ 2
3. Adjust The Grouping Coefficients: Although the equation is now BALANCED, the COEFFICIENTS must be WHOLE NUMBER INTEGERS. So we MULTIPLY the ENTIRE equation (per the laws of algebra), by the SMALLEST possible WHOLE NUMBER that ELIMINATES any FRACTIONS. In this case, we MULTIPLY by 2 to ELIMINATE the FRACTIONS for chlorine (πΆπ2 )
2 ππ + 1 πΆπ2 β 2 πππΆπ Made with by Prepineer | Prepineer.com
4. Verify Mass Balance Of The Atoms For Each Element: VERIFY that the NUMBER of ATOMS for each ELEMENT is the SAME on both sides of the CHEMICAL EQUATION:
π ππππ‘πππ‘π (2 ππ, 2 πΆπ) β πππππ’ππ‘π (2 ππ, 2 πΆπ)
CONCEPT EXAMPLE: Solid potassium nitrate decomposes when heated to produce solid potassium nitrate and oxygen gas. Given the skeleton equation below, the balance chemical equation for this reaction is most close to: πΎππ3 (π ) β πΎππ2 (π ) + π2 (π) A. πΎππ3 (π ) β 2 πΎππ2 (π ) + π2 (π) B. 2 πΎππ3 (π ) β 2 πΎππ2 (π ) + π2 (π) C. πΎππ3 (π ) β πΎππ2 (π ) + π2 (π) D. 2 πΎππ3 (π ) β 2 πΎππ2 (π ) + 2 π2 (π)
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SOLUTION: The GENERAL FORMULA for a CHEMICAL EQUATION is not directly provided to us in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. However, it is of upmost importance that we understand the fundamental concepts and applications revolving around this subject independent of the NCEES Supplied Reference Handbook. In this problem, we are GIVEN the SKELETON EQUATION and asked to BALANCE the EQUATION. As with any EQUATION in engineering, we can BALANCE CHEMICAL EQUATIONS using a simple 4 STEP PROCESS. 1. Write Out The Problem Statement:
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IDENTIFY all REACTANTS and PRODUCTS and WRITE their correct FORMULAS on each SIDE of the CHEMICAL EQUATION. Luckily, we are given the SKELETON EQUATION to work with, so we do not need to worry about TRANSLATING the NOMENCLATURE of the PROBLEM STATEMENT. For this STEP, we will just re-write the given SKELETON EQUATION: πΎππ3 (π ) β πΎππ2 (π ) + π2 (π)
2. Balance The Atoms For Each Element: Begin balancing the equation by trying different COEFFICIENTS to ADJUST the NUMBER of ATOMS of each ELEMENT to be the SAME on both sides of the CHEMICAL EQUATION. We can change the COEFFICIENTS (the numbers preceding the formulas), but not the SUBSCRIPTS (the number within chemical formulas), as changing the subscripts would change the IDENTITY of the SUBSTANCE. Looking at the number of ATOMS for potassium βKβ, we find there is 1 potassium atom in the REACTANTS, and 1 potassium atom in the PRODUCTS. Next, we look at the number of ATOMS for nitrogen βNβ, and find there is 1 nitrogen atom in the REACTANTS, and 1 nitrogen atom in the PRODUCTS.
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Lastly, we look at the number of ATOMS for oxygen βOβ, and find there is 3 oxygen atoms in the REACTANTS, and 4 oxygen atoms in the PRODUCTS. Therefore, we need to BALANCE the number of OXYGEN ATOMS on BOTH SIDE of the CHEMICAL EQUATION. As there is a SURPLUS of ONE OXYGEN ATOM in the PRODUCTS, we can use a COEFFICIENT of Β½ to REMOVE ONE OXYGEN ATOM, resulting in a NET SUM of 3 OXYGEN atoms on each SIDE of the EQUATION. 1 πΎππ3 (π ) β πΎππ2 (π ) + π2 2 (π) 3. Adjust The Grouping Coefficients: Although the equation is now BALANCED, the COEFFICIENTS must be WHOLE NUMBER INTEGERS. So we MULTIPLY the ENTIRE equation (per the laws of algebra), by the SMALLEST possible WHOLE NUMBER that ELIMINATES any FRACTIONS. In this case, we MULTIPLY by 2 to ELIMINATE the FRACTIONS for oxygen (π2 ) in the PRODUCTS. Re-writing the chemical reaction, we find:
2πΎππ3 (π ) β 2πΎππ2 (π ) + π2 (π) Made with
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4. Verify Mass Balance Of The Atoms For Each Element: VERIFY that the NUMBER of ATOMS for each ELEMENT is the SAME on both sides of the CHEMICAL EQUATION:
π ππππ‘πππ‘π (2 πΎ, 2 π, 6 π) β πππππ’ππ‘π (2 πΎ, 2 π, 6 π) As there is the same NUMBER of ATOMS for each SUBSTANCE on both sides of the CHEMICAL EQUATION, it is considered BALANCED.
Therefore, the correct answer choice is B. π ππππ (π¬) β π ππππ (π¬) + ππ (π )
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CONCEPT INTRO: A CHEMICAL REACTION is a PROCESS in which a SUBSTANCE or SUBSTANCES is CHANGED into one or more NEW SUBSTANCES. A CHEMICAL EQUATION represents the CHANGE that occurs during a CHEMICAL REACTION. A CHEMICAL EQUATION is a written REPRESENTATION of a CHEMICAL REACTION, using the SYMBOLS of each ELEMENT to indicate the AMOUNT of SUBSTANCE of each REACTANT and PRODUCT. An EXAMPLE of a simple CHEMICAL REACTION is the REACTION in which hydrogen gas (π»2 ) and oxygen gas (π2 ) COMBINE to product liquid water (π»2 0).
The GENERAL FORMULA for a CHEMICAL EQUATION is not directly provided to us in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Made with by Prepineer | Prepineer.com
However, it is of upmost importance that we understand the fundamental concepts and applications revolving around this subject independent of the NCEES Supplied Reference Handbook. The GENERAL FORM of a CHEMICAL REACTION demonstrates the component of a CHEMICAL PROCUESS by defining the REACTS that REACT and YIELD to produce PRODUCTS. π ππππ‘πππ‘π β πππππ’ππ‘π The YIELDS symbol " β " separates the REACTS from the PRODUCTS in a CHEMICAL EQUATION. A ONE-WAY YIELD symbol ββ " suggests that the REACTION occurs in only one direction. A TWO-WAY YIELDS symbol βββ indicates that the REACTION can occur REVERSIBLY, in both DIRECTIONS. The LEFT SIDE of the CHEMICAL EQUATION shows the REACTANTS, which are the ELEMENTS or COMPOUNDS that REACT in order for a CHEMICAL REACTION to OCCUR. The RIGHT SIDE of the CHEMICAL EQUATIONS shows the PRODUCTS, which are the ELEMENTS or COMPOUNDS that are PRODUCED as a RESULT of the CHEMICAL REACTION of the REACTANTS.
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In the earlier example, of hydrogen (π»2 ) gas and oxygen (π2 ) gas combining to form liquid water, HYDROGEN and OXYGEN would be the REACTANTS, and a WATER molecule would be the PRODUCT. 2π»2(π) + π2(π) β π»2 π(π) The STATE of MATTER of each SUBSTANCE is represented by an ABBREVIATION in the SUBSCRIPT of each REACTANT and PRODUCT. The NOTATE for each STATE of MATTER is as follows: β’ πΊππ = π(π) β’ πΏπππ’ππ = π(π) β’ πππππ = π(π ) β’ π΄ππ’πππ’π ππππ’π‘πππ = π(ππ) An AQUEOUS SOLUTION is a SOLID that is DISSOLVED in a LIQUID SOLUTION, resulting in a DILUTED SOLUTION. For example, given a sample of the oxygen that we breathe in as a gas, we would EXPRESS oxygen as: π2 (π) In a BALANCED CHEMICAL EQUATION, the NUMBER of ATOMS of each ELEMENT is EQUAL on both sides of the CHEMICAL EQUATION. The TOPIC of LAW OF CONSERVATION OF MASS is not directly provided to us in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Made with
by Prepineer | Prepineer.com
However, it is of upmost importance that we understand the fundamental concepts and applications revolving around this subject independent of the NCEES Supplied Reference Handbook. It is important to BALANCE CHEMICAL EQUATIONS in order to follow the LAW OF CONSERVATION MASS, which states that all the MASS or ATOMS present at the BEGINNING of a REACTION must be PRESENT in the FORM of a PRODUCT. Simply speaking, the LAW OF CONSERVATION states that there must be an EQUAL number of ATOMS of each ELEMENT in the REACTANTS as in the PRODUCTS of the CHEMICAL EQUATION. Before we BALANCE a CHEMICAL EQUATION, letβs look at the GENERAL FORM of the SKELETON EQUATION for any CHEMICAL EQUATION:
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β’ A, B, C, and D are the GROUPINGS, which represent the NUMBER of ATOMS in the REACTANT or PRODUCTS. A GROUPING is typically an ELEMENT, MOLECULE, or COMPOUND. β’ N, M, O, and P are the COEFFICIENTS of each GROUPING, which REPRESENT the NUMBER of MOLS for each ELEMENT, COMPOUND, or MOLECULE. β’ W, X, Y, and Z are the SUBSCRIPTS, which are part of the CHEMICAL FORMULA for a particular GROUPING and cannot be changed. When approaching a CHEMICAL EQUATION, it is important you understand the difference between COEFFICIENTS and SUBSCRIPTS. The COEFFICIENT is a WHOLE NUMBER INTEGER placed in front of an ELEMENT, COMPOUND, or MOLECULE, and denotes the AMOUNT of MOLES for a particular ELEMENT, COMPOUND, or MOLECULE. The SUBSCRIPT is placed behind the ELEMENT, MOLECULE, or COMPOUND and denotes the NUMBER of ATOMS in that GROUPING. To BALANCE an EQUATION, COEFFICIENTS are used to MODIFY the NUMBER of MOLES of REACTANTS and/or PRODUCT, such that MASS is CONSERVED on BOTH SIDE of the CHEMICAL EQUATION. CHANGING the COEFFICIENT changes the TOTAL NUMBER of MOLES of that ELEMENT or MOLECULE. The SUBSCRIPT, however cannot be changed, as altering a SUBSCRIPT would change the MOLECULE itself.
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The TOPIC of DIATOMIC MOLECULES is not directly provided to us in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. However, it is of upmost importance that we understand the fundamental concepts and applications revolving around this subject independent of the NCEES Supplied Reference Handbook. A DIATOMIC MOLECULE is a MOLECULE made of only TWO ATOMS, of either the SAME or different CHEMICAL ELEMENTS. Some ELEMENTS can only exist as DIATOMIC MOLECULES and are not physically able to exist as a single ATOM, but as a PAIR of ATOMS that share ELECTRONS to reach the nearest OCTET. The SEVEN ELEMENTS that exist only as DIATOMIC MOLECULES are hydrogen, nitrogen, oxygen, fluorine, chlorine, iodine, and bromine. You can easily remember the DIATOMIC ELEMENTS with the phrase:
πave πo π ear πf πce πold πeer Where: β’ H represents Hydrogen (π»2 ) β’ N represents Nitrogen (π2 ) β’ F represents Fluorine (πΉ2 ) β’ O represents Oxygen (π2 ) Made with
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β’ I represents Iodine (πΌ2 ) β’ C represents Chlorine (πΆπ2 ) β’ B represents Bromine (π΅π2 )
As with any EQUATION in engineering, we can BALANCE CHEMICAL EQUATIONS using a simple 4 STEP PROCESS. 1. Write Out The Problem Statement: IDENTIFY all REACTANTS and PRODUCTS and WRITE their correct FORMULAS on each SIDE of the CHEMICAL EQUATION. It is crucial to correctly TRANSLATE the PROBLEM STATEMENT to correctly WRITE out the CHEMICAL EQUATION. For example, a sample of sodium is mixed with chlorine gas to form sodium chloride. We can write the SKELETON EQUATION for this CHEMICAL REACTION as:
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ππ + πΆπ2 β πππΆπΏ
2. Balance The Atoms For Each Element: Begin balancing the equation by trying different COEFFICIENTS to ADJUST the NUMBER of ATOMS of each ELEMENT to be the SAME on both sides of the CHEMICAL EQUATION. We can change the COEFFICIENTS (the numbers preceding the formulas), but not the SUBSCRIPTS (the number within chemical formulas), as changing the subscripts would change the IDENTITY of the SUBSTANCE. 1 1 ππ + πΆπ2 β 1 ππ πΆπ 2
3. Adjust The Grouping Coefficients: Although the equation is now BALANCED, the COEFFICIENTS must be WHOLE NUMBER INTEGERS. So we MULTIPLY the ENTIRE equation (per the laws of algebra), by the SMALLEST possible WHOLE NUMBER that ELIMINATES any FRACTIONS. In this case, we MULTIPLY by 2 to ELIMINATE the FRACTIONS for chlorine (πΆπ2 )
2 ππ + 1 πΆπ2 β 2 πππΆπ Made with by Prepineer | Prepineer.com
4. Verify Mass Balance Of The Atoms For Each Element: VERIFY that the NUMBER of ATOMS for each ELEMENT is the SAME on both sides of the CHEMICAL EQUATION:
π ππππ‘πππ‘π (2 ππ, 2 πΆπ) β πππππ’ππ‘π (2 ππ, 2 πΆπ)
CONCEPT EXAMPLE: Solid potassium nitrate decomposes when heated to produce solid potassium nitrate and oxygen gas. Given the skeleton equation below, the balance chemical equation for this reaction is most close to: πΎππ3 (π ) β πΎππ2 (π ) + π2 (π) A. πΎππ3 (π ) β 2 πΎππ2 (π ) + π2 (π) B. 2 πΎππ3 (π ) β 2 πΎππ2 (π ) + π2 (π) C. πΎππ3 (π ) β πΎππ2 (π ) + π2 (π) D. 2 πΎππ3 (π ) β 2 πΎππ2 (π ) + 2 π2 (π)
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SOLUTION: The GENERAL FORMULA for a CHEMICAL EQUATION is not directly provided to us in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. However, it is of upmost importance that we understand the fundamental concepts and applications revolving around this subject independent of the NCEES Supplied Reference Handbook. In this problem, we are GIVEN the SKELETON EQUATION and asked to BALANCE the EQUATION. As with any EQUATION in engineering, we can BALANCE CHEMICAL EQUATIONS using a simple 4 STEP PROCESS. 1. Write Out The Problem Statement:
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IDENTIFY all REACTANTS and PRODUCTS and WRITE their correct FORMULAS on each SIDE of the CHEMICAL EQUATION. Luckily, we are given the SKELETON EQUATION to work with, so we do not need to worry about TRANSLATING the NOMENCLATURE of the PROBLEM STATEMENT. For this STEP, we will just re-write the given SKELETON EQUATION: πΎππ3 (π ) β πΎππ2 (π ) + π2 (π)
2. Balance The Atoms For Each Element: Begin balancing the equation by trying different COEFFICIENTS to ADJUST the NUMBER of ATOMS of each ELEMENT to be the SAME on both sides of the CHEMICAL EQUATION. We can change the COEFFICIENTS (the numbers preceding the formulas), but not the SUBSCRIPTS (the number within chemical formulas), as changing the subscripts would change the IDENTITY of the SUBSTANCE. Looking at the number of ATOMS for potassium βKβ, we find there is 1 potassium atom in the REACTANTS, and 1 potassium atom in the PRODUCTS. Next, we look at the number of ATOMS for nitrogen βNβ, and find there is 1 nitrogen atom in the REACTANTS, and 1 nitrogen atom in the PRODUCTS.
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Lastly, we look at the number of ATOMS for oxygen βOβ, and find there is 3 oxygen atoms in the REACTANTS, and 4 oxygen atoms in the PRODUCTS. Therefore, we need to BALANCE the number of OXYGEN ATOMS on BOTH SIDE of the CHEMICAL EQUATION. As there is a SURPLUS of ONE OXYGEN ATOM in the PRODUCTS, we can use a COEFFICIENT of Β½ to REMOVE ONE OXYGEN ATOM, resulting in a NET SUM of 3 OXYGEN atoms on each SIDE of the EQUATION. 1 πΎππ3 (π ) β πΎππ2 (π ) + π2 2 (π) 3. Adjust The Grouping Coefficients: Although the equation is now BALANCED, the COEFFICIENTS must be WHOLE NUMBER INTEGERS. So we MULTIPLY the ENTIRE equation (per the laws of algebra), by the SMALLEST possible WHOLE NUMBER that ELIMINATES any FRACTIONS. In this case, we MULTIPLY by 2 to ELIMINATE the FRACTIONS for oxygen (π2 ) in the PRODUCTS. Re-writing the chemical reaction, we find:
2πΎππ3 (π ) β 2πΎππ2 (π ) + π2 (π) Made with
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4. Verify Mass Balance Of The Atoms For Each Element: VERIFY that the NUMBER of ATOMS for each ELEMENT is the SAME on both sides of the CHEMICAL EQUATION:
π ππππ‘πππ‘π (2 πΎ, 2 π, 6 π) β πππππ’ππ‘π (2 πΎ, 2 π, 6 π) As there is the same NUMBER of ATOMS for each SUBSTANCE on both sides of the CHEMICAL EQUATION, it is considered BALANCED.
Therefore, the correct answer choice is B. π ππππ (π¬) β π ππππ (π¬) + ππ (π )
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