08 Chemical Equations Problem Set
CHEMICAL EQUATIONS | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: The balanced version of the chemical equation below is most close to: πΆ5 π»12 + π2 β πΆπ2 + π»2 π
A. πΆ5 π»12 + 8 π2 β 5 πΆπ2 + 6 π»2 π B. πΆ5 π»12 + π2 β 5 πΆπ2 + 6 π»2 π C. πΆ5 π»12 + 2 π2 β πΆπ2 + 6 π»2 π D. 5πΆ5 π»12 + 8 π2 β πΆπ2 + 6 π»2 π
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SOLUTION 1: 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: 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: πΆ5 π»12 + π2 β πΆπ2 + π»2 π
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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 carbon βCβ, we find there are 5 carbon atoms in the REACTANTS, and 1 carbon atom in the PRODUCTS. Next, we look at the number of ATOMS for hydrogen βHβ, and find there are 12 hydrogen atoms in the REACTANTS, and 2 hydrogen atoms in the PRODUCTS. Lastly, we look at the number of ATOMS for oxygen βOβ, and find there are 2 oxygen atoms in the REACTANTS, and 3 oxygen atoms in the PRODUCTS. Therefore, we need to BALANCE the number of carbon, hydrogen, and oxygen ATOMS on both SIDES of the CHEMICAL EQUATION. As there is a SURPLUS of 4 CARBON ATOMS in the REACTANTS, we can use a COEFFICIENT of 5 for the CARBON in the PRODUCTS, resulting in a NET SUM of 5 OXYGEN ATOMS on each SIDE of the EQUATION.
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Next, we look at the NUMBER of HYDROGEN ATOMS on each SIDE of the EQUATION, and find a SURPLUS of 10 HYDROGEN ATOMS in the PRODUCTS. We use a COEFFICIENT of 6 for the HYDROGEN ATOMS in the PRODUCTS, resulting in a NET SUM of 12 HYDROGEN ATOMS on each SIDE of the EQUATION. πΆ5 π»12 + π2 β 5 πΆπ2 + 6 π»2 π Lastly, we look at the NUMBER of OXYGEN ATOMS on each SIDE of the EQUATION, and find a SURPLUS of 14 OXYGEN ATOMS in the PRODUCTS. We use a COEFFICIENT of 8 for the OXYGEN ATOMS in the REACTANTS, resulting in a NET SUM of 16 OXYGEN ATOMS on each SIDE of the EQUATION. πΆ5 π»12 + 8 π2 β 5 πΆπ2 + 6 π»2 π 3. Adjust The Grouping Coefficients: Although the equation is now BALANCED, the COEFFICIENTS must be WHOLE NUMBER INTEGERS. As the COEFFICIENTS we used are all WHOLE NUMBER INTEGERS, we are good to go onto the FINAL STEP of verifying the BALANCE of ATOMS for each ELEMENT.
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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 (5 πΆ, 12 π», 16 π) β πππππ’ππ‘π (5 πΆ, 12 π», 16 π) As there is the same NUMBER of ATOMS for each SUBSTANCE on both sides of the CHEMICAL EQUATION, it is considered BALANCED. πΆ5 π»12 + 8 π2 β 5 πΆπ2 + 6 π»2 π
Therefore, the correct answer choice is A. πͺπ π―ππ + ππΆπ β ππͺπΆπ + ππ―π πΆ
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PROBLEM 2: The balanced version of the chemical equation below is most close to: πΉπ2 π3 (π ) + πΆ(π ) β πΉπ(π ) + πΆπ2 (π) A. 4πΉπ2 π3 (π ) + 2πΆ(π ) β πΉπ(π ) + 6πΆπ2 (π) B. πΉπ2 π3 (π ) + πΆ(π ) β 2πΉπ(π ) + 3πΆπ2 (π) C. 2πΉπ2 π3 (π ) + 3πΆ(π ) β 4πΉπ(π ) + 3πΆπ2 (π) D. 2πΆ3 π»4 + 4π2 β 5πΆπ2 + 2π»2 π
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SOLUTION 2: 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: 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: πΉπ2 π3 (π ) + πΆ(π ) β πΉπ(π ) + πΆπ2 (π)
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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 iron βFeβ, and we find there are 2 iron atoms in the REACTANTS, and 1 iron atom in the PRODUCTS. Next, we look at the number of ATOMS for oxygen βOβ, and find there are 3 oxygen atoms in the REACTANTS, and 2 oxygen atoms in the PRODUCTS. Lastly, we look at the number of ATOMS for carbon βCβ, and find there are 2 carbon atoms in the REACTANTS, and 1 carbon atom in the PRODUCTS. Therefore, we need to BALANCE the number of iron and oxygen ATOMS on both SIDES of the CHEMICAL EQUATION. As there is a SURPLUS of 1 IRON ATOM in the REACTANTS, we can use a COEFFICIENT of 2 for the IRON in the PRODUCTS, resulting in a NET SUM of 2 IRON ATOMS on each SIDE of the EQUATION. πΉπ2 π3 (π ) + πΆ(π ) β 2πΉπ(π ) + πΆπ2 (π) Made with
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Next, we look at the NUMBER of OXYGEN ATOMS on each SIDE of the EQUATION, and find a SURPLUS of 1 OXYGEN ATOM in the PRODUCTS. We use a COEFFICIENT of 3/2 for the OXYGEN ATOMS in the PRODUCTS, resulting a NET SUM of 3 OXYGEN ATOMS on each SIDE of the EQUATION.
3 πΉπ2 π3 (π ) + πΆ(π ) β 2πΉπ(π ) + πΆπ2 (π) 2 Lastly, we look at the NUMBER of CARBON ATOMS on each SIDE of the EQUATION, and find a SURPLUS of 3/2 CARBON ATOMS in the PRODUCTS. We use a COEFFICIENT of 3/2 for the CARBON ATOMS in the REACTANTS, resulting in a NET SUM of 3/2 CARBON ATOMS on each SIDE of the EQUATION. 3 3 πΉπ2 π3 (π ) + πΆ(π ) β 2πΉπ(π ) + πΆπ2 (π) 2 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.
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In this case, we MULTIPLY by 2 to ELIMINATE the FRACTIONS for CARBON (πΆ) and OXYGEN (π2 ) in the PRODUCTS and REACTANTS. Re-writing the chemical reaction, we find: 2πΉπ2 π3 (π ) + 3πΆ(π ) β 4πΉπ(π ) + 3πΆπ2 (π)
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 (4 πΉπ, 6 π, 3 πΆ) β πππππ’ππ‘π (4 πΉπ, 2 6 π, 6 πΆ) 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 C. ππ ππ ππ (π¬) + ππ(π¬) β ππ π(π¬) + ππππ (π )
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PROBLEM 3: The balanced version of the chemical equation below is most close to: π»2 π(π) + ππ2 (π) β π8 (π ) + π»2 π(π) A. 2π»2 π(π) + 4ππ2 (π) β π8 (π ) + 16π»2 π(π) B. 2π»2 π(π) + ππ2 (π) β 3π8 (π ) + 16π»2 π(π) C. π»2 π(π) + 3ππ2 (π) β 4π8 (π ) + 8π»2 π(π) D. 16π»2 π(π) + 8ππ2 (π) β 3π8 (π ) + 16π»2 π(π)
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SOLUTION 3: 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: 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: π»2 π(π) + ππ2 (π) β π8 (π ) + π»2 π(π)
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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 hydrogen βHβ, and find there are 2 hydrogen atoms in the REACTANTS, and 2 hydrogen atoms in the PRODUCTS. Next, we look at the number of ATOMS for sulfur βSβ, and find there is 1 sulfur atom in the REACTANTS, and 8 sulfur atoms in the PRODUCTS. Lastly, we look at the number of ATOMS for oxygen βOβ, and find there are 2 oxygen atoms in the REACTANTS, and 1 oxygen atom in the PRODUCTS. Therefore, we need to BALANCE the number of sulfur and oxygen ATOMS on both SIDES of the CHEMICAL EQUATION. As there is a SURPLUS of 1 OXYGEN ATOM in the REACTANTS, we can use a COEFFICIENT of 2 for the OXYGEN in the PRODUCT, resulting in a NET SUM of 2 OXYGEN ATOMS on each SIDE of the EQUATION. π»2 π(π) + ππ2 (π) β π8 (π ) + 2 π»2 π(π) Made with
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Next, we look at the NUMBER of HYDROGEN ATOMS on each SIDE of the EQUATION, and find a SURPLUS of 2 HYDROGEN ATOMS in the PRODUCTS. We use a COEFFICIENT of 2 for the HYDROGEN ATOMS in the REACTANTS, resulting a NET SUM of 4 OXYGEN ATOMS on each SIDE of the EQUATION.
2π»2 π(π) + ππ2 (π) β π8 (π ) + 2 π»2 π(π) Lastly, we look at the NUMBER of SULFUR ATOMS on each SIDE of the EQUATION, and find a SURPLUS of 5 SULFUR ATOMS in the PRODUCTS. We use a COEFFICIENT of 3/8 for the SULFUR ATOMS in the PRODUCTS, resulting in a NET SUM of 3CARBON ATOMS on each SIDE of the EQUATION.
2π»2 π(π) + ππ2 (π) β
3 π + 2 π»2 π(π) 8 8 (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 8 to ELIMINATE the FRACTIONS for SULFUR (π) in the PRODUCTS. Made with
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Re-writing the chemical reaction, we find: 16π»2 π(π) + 8ππ2 (π) β 3π8 (π ) + 16 π»2 π(π)
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 (32 π», 24 π, 16 π) β πππππ’ππ‘π (32 π», 24 π, 16 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 D. ππ ππ π(π ) + π πππ (π ) β π ππ (π¬) + ππ ππ π(π )
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PROBLEM 4: Which of the following elements is not diatomic? A. ππ₯π¦πππ B. πΌπππ C. πΌπππππ D. π΅ππππππ
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SOLUTION 4: 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
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Where: β’ H represents Hydrogen (π»2 ) β’ N represents Nitrogen (π2 ) β’ F represents Fluorine (πΉ2 ) β’ O represents Oxygen (π2 ) β’ I represents Iodine (πΌ2 ) β’ C represents Chlorine (πΆπ2 ) β’ B represents Bromine (π΅π2 ) As the only element not listed is iron βFEβ, we know that IRON is the only element listed that is not DIATOMIC. Therefore, the correct answer choice is B. Iron
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PROBLEM 5: Which of the following principles governs the balancing of chemical equations? A. π΄π£ππππππ β² π ππ’ππππ B. πΆππ’ππππ β² π πΏππ€ C. πΏππ€ ππ πΆπππ πππ£ππ‘πππ ππ ππππππ‘π’π D. πΏππ€ ππ πΆπππ πππ£ππ‘πππ ππ πππ π
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SOLUTION 5: 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. 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.
Therefore, the correct answer choice is D. Law of Conservation of Mass
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PROBLEM 1: The balanced version of the chemical equation below is most close to: πΆ5 π»12 + π2 β πΆπ2 + π»2 π
A. πΆ5 π»12 + 8 π2 β 5 πΆπ2 + 6 π»2 π B. πΆ5 π»12 + π2 β 5 πΆπ2 + 6 π»2 π C. πΆ5 π»12 + 2 π2 β πΆπ2 + 6 π»2 π D. 5πΆ5 π»12 + 8 π2 β πΆπ2 + 6 π»2 π
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SOLUTION 1: 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: 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: πΆ5 π»12 + π2 β πΆπ2 + π»2 π
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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 carbon βCβ, we find there are 5 carbon atoms in the REACTANTS, and 1 carbon atom in the PRODUCTS. Next, we look at the number of ATOMS for hydrogen βHβ, and find there are 12 hydrogen atoms in the REACTANTS, and 2 hydrogen atoms in the PRODUCTS. Lastly, we look at the number of ATOMS for oxygen βOβ, and find there are 2 oxygen atoms in the REACTANTS, and 3 oxygen atoms in the PRODUCTS. Therefore, we need to BALANCE the number of carbon, hydrogen, and oxygen ATOMS on both SIDES of the CHEMICAL EQUATION. As there is a SURPLUS of 4 CARBON ATOMS in the REACTANTS, we can use a COEFFICIENT of 5 for the CARBON in the PRODUCTS, resulting in a NET SUM of 5 OXYGEN ATOMS on each SIDE of the EQUATION.
πΆ5 π»12 + π2 β 5πΆπ2 + π»2 π Made with
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Next, we look at the NUMBER of HYDROGEN ATOMS on each SIDE of the EQUATION, and find a SURPLUS of 10 HYDROGEN ATOMS in the PRODUCTS. We use a COEFFICIENT of 6 for the HYDROGEN ATOMS in the PRODUCTS, resulting in a NET SUM of 12 HYDROGEN ATOMS on each SIDE of the EQUATION. πΆ5 π»12 + π2 β 5 πΆπ2 + 6 π»2 π Lastly, we look at the NUMBER of OXYGEN ATOMS on each SIDE of the EQUATION, and find a SURPLUS of 14 OXYGEN ATOMS in the PRODUCTS. We use a COEFFICIENT of 8 for the OXYGEN ATOMS in the REACTANTS, resulting in a NET SUM of 16 OXYGEN ATOMS on each SIDE of the EQUATION. πΆ5 π»12 + 8 π2 β 5 πΆπ2 + 6 π»2 π 3. Adjust The Grouping Coefficients: Although the equation is now BALANCED, the COEFFICIENTS must be WHOLE NUMBER INTEGERS. As the COEFFICIENTS we used are all WHOLE NUMBER INTEGERS, we are good to go onto the FINAL STEP of verifying the BALANCE of ATOMS for each ELEMENT.
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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 (5 πΆ, 12 π», 16 π) β πππππ’ππ‘π (5 πΆ, 12 π», 16 π) As there is the same NUMBER of ATOMS for each SUBSTANCE on both sides of the CHEMICAL EQUATION, it is considered BALANCED. πΆ5 π»12 + 8 π2 β 5 πΆπ2 + 6 π»2 π
Therefore, the correct answer choice is A. πͺπ π―ππ + ππΆπ β ππͺπΆπ + ππ―π πΆ
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PROBLEM 2: The balanced version of the chemical equation below is most close to: πΉπ2 π3 (π ) + πΆ(π ) β πΉπ(π ) + πΆπ2 (π) A. 4πΉπ2 π3 (π ) + 2πΆ(π ) β πΉπ(π ) + 6πΆπ2 (π) B. πΉπ2 π3 (π ) + πΆ(π ) β 2πΉπ(π ) + 3πΆπ2 (π) C. 2πΉπ2 π3 (π ) + 3πΆ(π ) β 4πΉπ(π ) + 3πΆπ2 (π) D. 2πΆ3 π»4 + 4π2 β 5πΆπ2 + 2π»2 π
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SOLUTION 2: 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: 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: πΉπ2 π3 (π ) + πΆ(π ) β πΉπ(π ) + πΆπ2 (π)
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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 iron βFeβ, and we find there are 2 iron atoms in the REACTANTS, and 1 iron atom in the PRODUCTS. Next, we look at the number of ATOMS for oxygen βOβ, and find there are 3 oxygen atoms in the REACTANTS, and 2 oxygen atoms in the PRODUCTS. Lastly, we look at the number of ATOMS for carbon βCβ, and find there are 2 carbon atoms in the REACTANTS, and 1 carbon atom in the PRODUCTS. Therefore, we need to BALANCE the number of iron and oxygen ATOMS on both SIDES of the CHEMICAL EQUATION. As there is a SURPLUS of 1 IRON ATOM in the REACTANTS, we can use a COEFFICIENT of 2 for the IRON in the PRODUCTS, resulting in a NET SUM of 2 IRON ATOMS on each SIDE of the EQUATION. πΉπ2 π3 (π ) + πΆ(π ) β 2πΉπ(π ) + πΆπ2 (π) Made with
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Next, we look at the NUMBER of OXYGEN ATOMS on each SIDE of the EQUATION, and find a SURPLUS of 1 OXYGEN ATOM in the PRODUCTS. We use a COEFFICIENT of 3/2 for the OXYGEN ATOMS in the PRODUCTS, resulting a NET SUM of 3 OXYGEN ATOMS on each SIDE of the EQUATION.
3 πΉπ2 π3 (π ) + πΆ(π ) β 2πΉπ(π ) + πΆπ2 (π) 2 Lastly, we look at the NUMBER of CARBON ATOMS on each SIDE of the EQUATION, and find a SURPLUS of 3/2 CARBON ATOMS in the PRODUCTS. We use a COEFFICIENT of 3/2 for the CARBON ATOMS in the REACTANTS, resulting in a NET SUM of 3/2 CARBON ATOMS on each SIDE of the EQUATION. 3 3 πΉπ2 π3 (π ) + πΆ(π ) β 2πΉπ(π ) + πΆπ2 (π) 2 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.
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In this case, we MULTIPLY by 2 to ELIMINATE the FRACTIONS for CARBON (πΆ) and OXYGEN (π2 ) in the PRODUCTS and REACTANTS. Re-writing the chemical reaction, we find: 2πΉπ2 π3 (π ) + 3πΆ(π ) β 4πΉπ(π ) + 3πΆπ2 (π)
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 (4 πΉπ, 6 π, 3 πΆ) β πππππ’ππ‘π (4 πΉπ, 2 6 π, 6 πΆ) 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 C. ππ ππ ππ (π¬) + ππ(π¬) β ππ π(π¬) + ππππ (π )
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PROBLEM 3: The balanced version of the chemical equation below is most close to: π»2 π(π) + ππ2 (π) β π8 (π ) + π»2 π(π) A. 2π»2 π(π) + 4ππ2 (π) β π8 (π ) + 16π»2 π(π) B. 2π»2 π(π) + ππ2 (π) β 3π8 (π ) + 16π»2 π(π) C. π»2 π(π) + 3ππ2 (π) β 4π8 (π ) + 8π»2 π(π) D. 16π»2 π(π) + 8ππ2 (π) β 3π8 (π ) + 16π»2 π(π)
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SOLUTION 3: 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: 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: π»2 π(π) + ππ2 (π) β π8 (π ) + π»2 π(π)
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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 hydrogen βHβ, and find there are 2 hydrogen atoms in the REACTANTS, and 2 hydrogen atoms in the PRODUCTS. Next, we look at the number of ATOMS for sulfur βSβ, and find there is 1 sulfur atom in the REACTANTS, and 8 sulfur atoms in the PRODUCTS. Lastly, we look at the number of ATOMS for oxygen βOβ, and find there are 2 oxygen atoms in the REACTANTS, and 1 oxygen atom in the PRODUCTS. Therefore, we need to BALANCE the number of sulfur and oxygen ATOMS on both SIDES of the CHEMICAL EQUATION. As there is a SURPLUS of 1 OXYGEN ATOM in the REACTANTS, we can use a COEFFICIENT of 2 for the OXYGEN in the PRODUCT, resulting in a NET SUM of 2 OXYGEN ATOMS on each SIDE of the EQUATION. π»2 π(π) + ππ2 (π) β π8 (π ) + 2 π»2 π(π) Made with
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Next, we look at the NUMBER of HYDROGEN ATOMS on each SIDE of the EQUATION, and find a SURPLUS of 2 HYDROGEN ATOMS in the PRODUCTS. We use a COEFFICIENT of 2 for the HYDROGEN ATOMS in the REACTANTS, resulting a NET SUM of 4 OXYGEN ATOMS on each SIDE of the EQUATION.
2π»2 π(π) + ππ2 (π) β π8 (π ) + 2 π»2 π(π) Lastly, we look at the NUMBER of SULFUR ATOMS on each SIDE of the EQUATION, and find a SURPLUS of 5 SULFUR ATOMS in the PRODUCTS. We use a COEFFICIENT of 3/8 for the SULFUR ATOMS in the PRODUCTS, resulting in a NET SUM of 3CARBON ATOMS on each SIDE of the EQUATION.
2π»2 π(π) + ππ2 (π) β
3 π + 2 π»2 π(π) 8 8 (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 8 to ELIMINATE the FRACTIONS for SULFUR (π) in the PRODUCTS. Made with
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Re-writing the chemical reaction, we find: 16π»2 π(π) + 8ππ2 (π) β 3π8 (π ) + 16 π»2 π(π)
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 (32 π», 24 π, 16 π) β πππππ’ππ‘π (32 π», 24 π, 16 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 D. ππ ππ π(π ) + π πππ (π ) β π ππ (π¬) + ππ ππ π(π )
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PROBLEM 4: Which of the following elements is not diatomic? A. ππ₯π¦πππ B. πΌπππ C. πΌπππππ D. π΅ππππππ
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SOLUTION 4: 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
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Where: β’ H represents Hydrogen (π»2 ) β’ N represents Nitrogen (π2 ) β’ F represents Fluorine (πΉ2 ) β’ O represents Oxygen (π2 ) β’ I represents Iodine (πΌ2 ) β’ C represents Chlorine (πΆπ2 ) β’ B represents Bromine (π΅π2 ) As the only element not listed is iron βFEβ, we know that IRON is the only element listed that is not DIATOMIC. Therefore, the correct answer choice is B. Iron
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PROBLEM 5: Which of the following principles governs the balancing of chemical equations? A. π΄π£ππππππ β² π ππ’ππππ B. πΆππ’ππππ β² π πΏππ€ C. πΏππ€ ππ πΆπππ πππ£ππ‘πππ ππ ππππππ‘π’π D. πΏππ€ ππ πΆπππ πππ£ππ‘πππ ππ πππ π
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SOLUTION 5: 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. 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.
Therefore, the correct answer choice is D. Law of Conservation of Mass
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