MISC 476 Interpreting and Balancing Chemical Equations

Transcription

1 MISC 476 Interpreting and Balancing Chemical Equations Prepared by M. L. Gillette, Indiana University Kokomo and H. A. Neidig, Lebanon Valley College Purpose of the Experiment: Interpret and balance chemical equations. Background Information Imagine models of an apartment building and a recreation center, built out of toy blocks. Suppose that we disassembled those structures and used the same blocks to construct models of three houses and two garages, with no blocks left over. We could summarize the process using the expression shown in Equation 1. apartment building + recreation center 3 houses + 2 garages (Eq. 1) The process of starting with one arrangement of building blocks, disassembling the arrangement, and reassembling the blocks into a different arrangement is like the process that occurs during a chemical reaction. We can represent a chemical reaction by writing an expression called a chemical equation, as shown in Equation 2. reactants products (Eq. 2) In chemical equations, we represent the starting arrangements, or structures, called the reactants, and the final structures, called the products, using chemical formulas. We indicate the rearrangement process using an arrow ( ), which we read as either "yields" or "produces." Interpreting Chemical Equations If we heat metallic nickel (Ni), a silvery solid, with antimony (Sb), a silvery-white solid, we produce nickel antimonide (NiSb), a light copper-red solid. We can describe this process using the chemical equation shown in Equation 3. Ni(s, silver) + Sb(s, silver-white) heat ""# NiSb(s, copper-red) (Eq. 3) In Equation 3, the information in parentheses following the elemental symbols and formulas indicates the physical states (solid, liquid, or gas) and colors of the reactants and products. The "s" stands for solid, "g" for gas, and "l" for liquid. If the substance is dissolved in water, we write "aq," which stands for aqueous. Frequently, we omit the color descriptions in order to make the notation simpler. The information above the reaction arrow describes reaction condition's, such as the application of heat. Write the chemical equations for each of the descriptions in Problem Set 1. Balancing Equations The disassembly and reassembly process represented by Equation 1 illustrates a principle called the conservation of matter: no blocks were added to make the product

2 structures, and none were left over. When substances are combined in the correct proportions, based on the appropriate chemical equation, every reactant atom should be incorporated into a product; no extra atoms should be added, and none should be left over. We establish the correct proportions for a reaction by balancing the equation for the reaction. Balancing equations is a bookkeeping procedure that helps us in counting the number of atoms, molecules, and moles involved in a chemical reaction. First, we need to recognize the number of atoms of each element that are in a unit of a compound. Potassium telluride has the formula K 2 Te. A subscript after a symbol indicates how many of that atom are in a unit of the compound. The subscript after the symbol K indicates that there are two K atoms in one unit of K 2 Te. There is no subscript for Te. It is understood that if there is no subscript after the symbol for an atom the subscript is one. Hence, there is only one atom of Te in a unit of K 2 Te. In a unit of potassium disulfide (K 2 S 4 ). there are two atoms of potassium and four atoms of sulfur. In a molecule or unit of cyclohexane (C 6 H 12 ), there are six carbon atoms and twelve hydrogen atoms. In the formula for some compounds, we find that a group of atoms is enclosed in parentheses and that there is a subscript following the parenthesis. An example is radium iodate, Ra(IO 3 ) 2. The subscript three after the oxygen indicates that in the iodate ion (IO 3 - ) there are three oxygen atoms. However, the subscript two appearing after the parentheses indicates that there are two iodate ions in the compound. Therefore, there are two iodine atoms in a unit of Ra(IO 3 ) 2 while the number of oxygen atoms in a unit of Ra(IO 3 ) 2 is six. Indicate the number of atoms in a unit of each of the compounds in Problem Set 2. Sometimes, an equation is already balanced when it is written in its simplest form. For example, Equation 3 is already balanced because one Ni atom and one Sb atom are on each side of the equation. In most cases, however, we have to do additional work to balance an equation. A chemical equation is quantitatively useful only when it is properly balanced. eonsider the equation for the reaction of sodium (Na) with Sb to produce sodium antimonide (Na3Sb), shown in Equation 4. Na(s, silver) + Sb(s, silver-white) Na 3 Sb(s, black) (Eq. 4) Equation 4 is not balanced. One Na atom and one Sb atom are on the reactant side, the left side of the equation. Three Na atoms and one Sb atom are on the product side, the right side of the equation. Due to the conservation of matter, it is not possible to have one atom of Na as the reactant and three atoms of Na in the product. We need to balance the equation by determining the correct proportions of reactants and products. If more than one of any species are involved in a chemical reaction, we write the number involved immediately preceding the appropriate symbol or formula in the equation for that reaction. The numbers preceding the symbols or formulas in a chemical equation are the coefficients of the equation. We insert the appropriate coefficients to balance chemical equations. When no coefficient precedes the symbol or formula for a species, a coefficient of 1 is understood to be there. Thus, the coefficients in Equation 1 are 1, 1, 3, and 2, respectively. To balance Equation 4, we write a coefficient of 3 in front of Na, as shown in Equation 4a.

3 3 Na(s, silver) + Sb(s, silver-white) Na 3 Sb(s, black) (Eq. 4a) Now there are three Na atoms and one Sb atom on each side of the equation. Therefore, the equation is balanced. It is essential that we do not alter the chemical formula of any reactant or product species in the process of balancing an equation. For example, if we change the subscripts in the chemical formula for a species, we alter the identity of that reactant or product species. The resulting equation may be balanced, but it will represent a different chemical transformation from the one represented by the original (unbalanced) equation. We can interpret coefficients in chemical equations in either of two ways: as the number of individual units involved in a reaction, or as the number of moles of those units involved in the reaction. Thus, based on Equation 4a, we can say that three Na atoms react with one Sb atom to produce one unit of Na 3 Sb, or we can say that 3 mol of Na atoms react with 1 mol of Sb atoms to produce 1 mol of Na 3 Sb units. When we write a chemical equation, we use chemical symbols to represent the elements and compounds involved in the form in which they participate in the reaction. For example, when they are in their elemental, or chemically uncombined forms, metal atoms are not written as if they are bonded to each other, so they can be considered as essentially reacting individually. Therefore, we can represent them in a chemical equation by simply using the appropriate chemical symbol, as shown in Equation 4a for Na and Sb. In contrast, some nonmetallic elements exist naturally not as individual atoms, but as molecules composed of two or more atoms of the element. Molecules are groupings of atoms held together by shared electron pairs. For example, hydrogen, nitrogen, oxygen, and the halogens (fluorine, chlorine, bromine, and iodine) occur as diatomic (two atom) molecules: H 2, N 2, O 2, F 2, Cl 2, Br 2, and I 2 respectively. We refer to H 2 as molecular hydrogen and H as an atom of hydrogen. Similarly, phosphorus occurs naturally as P 4, and sulfur occurs as S 8. These are the formulas that we use in chemical equations representing reactions involving these molecular elements. For example, carbon (C) reads with molecular oxygen (O 2 ) to form carbon(iv) dioxide (CO 2 ), as shown in Equation 5, which is balanced. C (s) + O 2 (g) CO 2 (g) (Eq. 5) Another illustration of the correct use of elemental formulas is the reaction of P 4 with O 2 to form tetraphosphorus decaoxide (P 4 O 10 ), shown in Equation 6, which is balanced. P 4 (s) + 5 O 2 (g) P 4 O 10 (s) (Eq.6) Notice that five O 2 molecules react with one P 4 molecule to form one P 4 O 10 molecule. Hydrogen (H 2 ) and nitrogen (N 2 ) gases react to form ammonia (NH 3 ) gas. The balanced equation for this reaction is shown in Equation 7. 3 H 2 (g) + N 2 (g) 2 NH 3 (g) (Eq. 7)

4 As previously discussed, we can interpret Equation 7 in either of two ways. We can say that three H 2 molecules (a total of six H atoms) react with one N 2 molecule (two N atoms). Two NH 3 molecules are produced, containing a total of two N atoms and six H atoms. Or, we can say that 3 mol of H 2 molecules (6 mol of hydrogen atoms) react with 1 mol of N 2 molecules (2 mol of N atoms). Two moles of NH 3 molecules are produced, containing a total of 2 mol of N atoms and 6 mol of H atoms. There are other sets of coefficients that we could use to balance Equation 7. For example, coefficients of 6, 2, and 4, respectively, also work, as shown in Equation 7a. 6 H 2 (g) + 2 N 2 (g) 4 NH 3 (g) (Eq. 7a) To simplify equations as much as possible, we choose the smallest whole number coefficients we can. Because we can divide 6, 2, and 4 by 2 and yield whole numbers, we prefer to write the equation in the form shown in Equation 7. Balance the equations in Problem Sets 3 and 4. Problem Set 1 Write a chemical equation for each of the following descriptions, indicating the physical states of the reactants and products. 1. If we react metallic germanium (Ge), a gray solid, with bromine, a red-brown liquid, we produce germanium(li) bromide (GeBr 2 ). a colorless solid. 2. If we heat metallic iron (Fe), a silvery solid, with metallic selenium (Se), a gray solid, we produce iron(ll) selenide (FeSe), a black solid. 3. When we heat metallic magnesium (Mg), a silvery solid, with hydrogen (H 2 ), a colorless gas, we produce magnesium hydride (MgH 2 ), a white solid. 4. When we react metallic copper (Cu), a red-brown solid, with fluorine (F 2 ), a yellowgreen gas, we produce copper(li) fluoride (CuF 2 ), a white solid.

5 5. When we heat magnesium carnonate (MgCO 3 ), a white solid, we produce magnesium oxide (MgO), a white solid, and carbon(lv) dioxide (CO 2 ), a colorless gas. Problem Set 2 Indicate the number of atoms of each element that are in one unit of the following compounds. 1. CaO Ca O 2. Na 2 S Na S 3. FeBr 3 Fe Br 4. Co 2 O 3 Co O 5. Cu 3 Se 2 Cu Se 6. K 2 SO 4 K S O 7. FeC 2 O 4 Fe C O 8. K 2 Cr 2 O 7 K Cr O 9. Mg(NO 3 ) 2 Mg N O 10. Al 2 (SO 3 ) 3 Al S O Problem Set 3 1. Rewrite and balance each of the following chemical equations in the space provided. a. Na (s) + Cl 2 (g) NaCl (s)

6 b. Mg (s) + Cl 2 (g) MgCl 2 (s) c. H 2 (g) + Cl 2 (g) HCl (g) d. H 2 (g) + O 2 (g) H 2 O 2 (l) e. H 2 O (l) H 2 (g) + O 2 (g) f. Bi (s) + F 2 (g) BiF 3 (s) g. Al (s) + O 2 (s) Al 2 O 3 (s)

7 h. Al (s) + Cl 2 (g) AlCl 3 (s) i. CaCO 3 (s) + CaO (s) CO 2 (g) j. KClO 3 (s) KCl (s) + O 2 (g) k. N 2 O 5 (g) NO 2 (g) + O 2 (g) l. P 4 O 10 (s) + H 2 O (l) H 3 PO 4 (s) m. Fe 2 O 3 (s) + Al (s) Fe (s) + Al 2 O 3 (s)

8 n. Sb 2 S 3 (s) + Fe (s) Sb (s) + FeS (s) o. NH 3 (g) + SO 2 (g) NO (g) + S (s) + H 2 O (g) p. NH 3 (g) + O 2 (g) NO (g) + H 2 O (g) 2. What is the sum of the coefficients used in the balanced form of the following equations? Example: The sum of the coefficients for 2 H 2 + O 2 2 H 2 O is 5 a. In Question 1.a. b. In Question 1.o. Problem Set 4 For each of the following incorrectly written equations, write the correct balanced equation, using the appropriate coefficients. Do not change any of the subscripts. 1. Fe + Br 2 2 FeBr H I 2 4 HI

9 3. 2 Ni + 4 CO 2 Ni(CO) K + 2 N 2 3 KN KClO 3 4 KCl + 3 O 2 6. Si + 4 F 2 SiF 4