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Objective Four: The student will demonstrate an understanding of the structures and properties of matter. TAKS Objective: (4) 8C The student knows that changes in matter affect everyday life. The student is expected to investigate and identify the law of conservation of mass. TEKS: IPC 8C: he student is expected to investigate and identify the law of conservation of mass. Objectives: 1. 2. 3. The student will need to be able to apply the law of conservation of mass. The student will need to be able to balance simply chemical equations. The student will need to be able to correctly identify a balanced chemical equation. Critical Attributes of the Concept: The teacher can put the following up on the board: I + O + P + C TR and have the students try to figure out what you are trying to communicate. Then put the following numbers in the places: 4 I + 3 O + 1 P + 1 C 9 TR. If the students haven’t figured it out when the numbers are put in place, give then the hint “sports”. The “equation” describes the Texas Rangers – 4 infielders, 3 outfielders, 1 pitcher, and 1 catcher. Have the students make up their own “equations” (examples 2 A + 2 L + 1 T B; 2 arms + 2 legs + 1 trunk = 1 body –or—have then describe what makes up a car). Once the students have the concept of equations, make sure that they know that they must be “balanced”. All chemical equations must satisfy the Law of Conservation of Mass, i.e. the mass of the reactants must equal the mass of the products. Remember that chemical equations are written: REACTANTS PRODUCTS Critical Attributes of the Skill: The following are the rules for writing balance chemical equations: 1. Write the formula of all reactants on the left of the arrow and products are written on the right. 2. Make sure that all formulas are written correctly and know that formulas can NOT be changed when you balance an equation. 3. Count the number of times each atom appears on (a) the reactant side, and (b) the product side of an equation. 4. If the numbers for (a) and for (b) agree, the equation is balanced. 5. If they do not agree, the equation is NOT balanced. An equation can be balanced by using coefficients. a. Coefficients are whole numbers placed before a chemical formula to show how many of each atom is present. b. Coefficients are used to “multiple” the number of each atom present. c. An example is 2 (NH4)2SO4 which means that you have: i. 4 nitrogen (N) ii. 16 hydrogen (H) iii. 2 sulfur (S) iv. 8 oxygen (O) d. When no coefficient is shown, it is assumed to be one. See the enclosed sheet titled Balancing Chemical Equations for a step by step guided practice for the student. Answers to the two practice problems are: 2 Al + N2 2 AlN 2 NaCl + H2SO4 Na2SO4 + 2 HCl Answers to Worksheets: Worksheet #1 1. B 2. C 3. A 4. C 5. C 6. C 7. B 8. C 9. C Law of Conservation of Mass is also called Law of Conservation of Matter. All equations must be balanced so that they show that the Law of Conservation of Mass is satisfied. The equation is already balanced, nothing needs to be done with it. The balanced equation is SnS2 + 6 HCl H2SnCl6 + 2 H2S The balanced equation is 3 Br2 + 2 AlCl3 2 AlBr3 + 3 Cl2 The equation that the problem tell us is: K + O2 K2O. Since we have 3.91 grams of K and 7.11 grams of K2O, and the mass of the products (K2O) must equal the mass of the reactants (K + O2), simply subtract the mass of K from the mass of products to find the oxygen. 7.11 grams – 3.91 grams = 3.20 grams Since the three is the subscript on the ( ) – it means multiple the elements on the inside of the ( ). The balanced equation is: 3 RbOH + H3PO4 Rb3PO4 + 3 H2O. The equation in the problem is S + F2 SF6. For an explanation of how to solve the problem, see number six. The answer is 5.722 grams – 1.256 grams = 4.466 grams Worksheet #2 1. A 2. E 3. B 4. D 5. E 6. C 7. C 8. B Atoms are rearranged in when they make products…they are never destroyed. The balance equation is: P4 + 10 Cl2 4 PCl5. Remember that there is a 1 when there is no coefficient. Law of Conservation of Mass = mass of products and reactants must be equal. The balanced equation is: 2 C9H18 + 27 O2 18 CO2 + 18 H2O. The balanced equation is: 2 C4H10 + 13 O2 CO2 + 10 H2O. Since the three is the subscript on the ( ) – it means multiple the elements on the inside of the ( ). The mass of the products (AgCl and NaNO3) must be equal to the mass of the reactants (NaCl + AgNO3). So, 143 + 85 = 58 + X X = 170. The balanced equation is: 2 Cr + 3 H2SO4 Cr2(SO4)3 + 3 H2. Balancing Chemical Equations Compounds are formed from rearrangements and new combinations of atoms. These rearrangements are represented by BALANCED EQUATIONS. Coefficients are used to balance chemical equations. Remember, Subscripts are used to write formulas. Coefficients are used to balance equations. The following equation is for the formation of a basketball team. unbalanced: balanced: GUARDS + FORWARDS + CENTER ----- G2F2C 2GUARDS + 2FORWARDS + CENTER ----- G2F2C When a chemical equation is balanced, EACH SIDE OF THE EQUATION HAS THE SAME NUMBER OF EACH ELEMENT. When hydrogen and oxygen react, water is the product. The chemical equation for this reaction is: H2 + O2 H2O This equation is not balanced, since the left hand side has two oxygen atoms, while the right hand side has only one. The following equation is the proper balanced equation. 2H2 + O2 2H2O There is a simple process to follow to balance chemical equations. The following example will illustrate the process. KCl + O2 ----- KClO3 Step 1: Do a preliminary atom count. ELEMENT K Cl O LEFT SIDE 1 1 2 RIGHT SIDE 1 1 3 Step 2: To make the number of oxygen atoms equal on both sides, place a COEFFICIENT of 2 before the KClO3 and a COEFFICIENT of 3 before the O2. This will give 6 oxygen atoms on each side of the equation. KCl + 3O2 ----- 2KClO3 Do another atom count on both sides. ELEMENT K Cl O LEFT SIDE RIGHT SIDE 1 1 2 6 1 2 1 2 3 6 Our equation is not yet balanced because now the number of potassium and chlorine atoms differ. Step 3: To balance the number of potassium and chlorine atoms place a coefficient of 2 before the KCl. 2KCl + 3O2 ----- 2KClO3 Do another atom count on both sides. ELEMENT K Cl O LEFT SIDE RIGHT SIDE 1 2 1 2 2 6 1 2 1 2 3 6 The equation is now balanced. Here is another example for you to fill in. Balance the following equation: Al + O2 Al2O3 Step 1: Do a preliminary atom count. ELEMENT LEFT SIDE RIGHT SIDE Al O Step 2: To make the number of oxygen atoms equal on both sides, place a coefficient of 2 before the Al2O3 and a coefficient of 3 before the O2. This will give 6 oxygen atoms on each side of the equation. Al + ___ O2 ____ Do another atom count on both sides. Al2O3 ELEMENT LEFT SIDE RIGHT SIDE Al O Our equation is not yet balanced because now the number of aluminum atoms differ. Step 3: To balance the number of aluminum atoms place a coefficient of ___ Al + ____ O2 ______ before the Al. ___ Al2O3 Do another atom count on both sides. ELEMENT LEFT SIDE RIGHT SIDE Al O The equation is now balanced. Do the following problems. Al + N2 AlN NaCl + H2SO4 Na2SO4 + HCl Name ____________________________________ Period _______________ Worksheet #1 1. What scientific (natural) law serves as the basis for balancing chemical equations by requiring that there be no observable change in the quantity of matter during a chemical reaction? a) Law of Conservation of Energy b) Law of Multiple Proportions c) Law of Conservation of Matter d) Law of Definite Proportions 2. To satisfy the law of conservation of mass, a chemical equation must a) be written in words b) have subscripts c) be balanced d) show the physical state of each of the reactants and products e) have a catalyst 3. What is the coefficient for CO when the following equation is balanced with the smallest whole number coefficients (space has been provided for you to do your work)? CO FeO + CO2 + Fe a) 1 b) 3 c) 4 d) 5 4. Balance the following equation with the smallest whole number coefficients. What is the coefficient for HCl in the balanced equation (space has been provided for you to do your work)? SnS2 + HCl H2SnCl6 + H2 S a) 3 b) 4 c) 6 d) 12 5. What is the coefficient for Br2 when the following equation is balanced with the smallest whole number coefficients (space has been provided for you to do your work)? Br2 + a) 1 b) 2 c) 3 d) 4 AlCl3 AlBr3 + Cl2 6. 3.91 g of potassium metal is allowed to react with oxygen, O2. The reaction yields 7.11 g of K2O. How much oxygen was reacted? a) b) c) d) 7. 0.10 grams 1.60 grams 3.20 grams 11.02 grams How many nitrogen atoms are represented by the following chemical formula, (NH4)3PO4? a) 1 b) 3 c) 12 d) 15 e) 24 8. Balance the following equation with the smallest whole number coefficients. Choose the answer that is the sum of the coefficients in the balanced equation. Space has been provided for you to do your work. RbOH + H3PO4 Rb3PO4 + H2 O a) 4 b) 6 c) 8 d) 10 e) 12 9. 1.256 g of sulfur is combined with fluorine, F2, to produce the 5.722 g of the compound SF6. How much fluorine was used in the reaction? a) b) c) d) 1.954 grams 3.210 grams 4.466 grams 6.978grams Name _______________________________________ Period _________________ Worksheet #2 1. In a chemical reaction, a) atoms may be rearranged b) atoms may be destroyed c) atoms may be created d) atoms must turn into molecules e) atoms must change their subscripts 2. Balance the following equation with the smallest whole number coefficients. Choose the answer that is the sum of the coefficients in the balanced equation. Space has been provided for you to do your work. P4 Cl2 + PCl5 a) 7 b) 9 c) 11 d) 13 e) 15 3. In a chemical reaction, the total mass of the reactants a) is greater than the total mass of the products b) is equal to the mass of the products c) is less than the total mass of the products d) is equal to the mass of the catalyst e) can not be determined 4. Balance the following equation with the smallest whole number coefficients. What is the coefficient for O2 in the balanced equation (space has been provided for you to do your work)? C9H18 + O2 CO2 + H2 O a) 5 b) 9 c) 14 d) 27 e) 54 5. Balance the following equation with the smallest whole number coefficients. Choose the answer that is the sum of the coefficients in the balanced equation. Space has been provided for you to do your work. C4H10 + a) 23 b) 27 c) 29 d) 30 e) 33 O2 CO2 + H2 O 6. The number of atoms of oxygen present in Al(NO3)3 is a) 3 b) 6 c) 9 d) 13 e) 15 7. Knowing that 58 grams of NaCl is used to produce 143 grams of AgCl and 85 grams of NaNO3, much AgNO3 is used in the following reaction? NaCl + AgNO3 AgCl + NaNO3 a) 58 b) 116 c) 170 d) 286 e) 398 8. Balance the following equation with the smallest whole number coefficients. Choose the answer that is the sum of the coefficients in the balanced equation. Space has been provided for you to do your work. Cr a) 7 b) 9 c) 11 d) 13 e) 15 + H2SO4 Cr2(SO4)3 + H2