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Balancing Chemical Equations

Balancing Chemical Equations. Key Terms: Balanced Chemical Equations Coefficient Subscript. Lesson Objectives. The student will: explain the roles of subscripts and coefficients in chemical equations.

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Balancing Chemical Equations

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  1. Balancing Chemical Equations Key Terms: Balanced Chemical Equations Coefficient Subscript

  2. Lesson Objectives • The student will: • explain the roles of subscripts and coefficients in chemical equations. • write a balanced chemical equation when given the unbalanced equation for any chemical reaction. • explain the role of the law of conservation of mass in a chemical reaction.

  3. Vocabulary • balanced chemical equation • a chemical equation in which the number of each type of atom is equal on the two sides of the equation • Coefficient • a whole number that appears in front of a formula in a balanced chemical equation • subscript • part of the chemical formula that indicates the number of atoms of the preceding element

  4. Introduction • Even though chemical compounds are broken up to form new compounds during a chemical reaction, atoms in the reactants do not disappear, nor do new atoms appear to form the products. In chemical reactions, atoms are never created or destroyed. The same atoms that were present in the reactants are present in the products. The atoms are merely re-organized into different arrangements. In a complete chemical equation, the two sides of the equation must be balanced. That is, in a balanced chemical equation, the same number of each atom must be present on the reactant and product sides of the equation.

  5. Balancing Equations • The process of writing a balanced chemical equation involves three steps. • Step 1: Know what the reactants and products are, and write a word equation for the reaction. • Step 2: Write the formulas for all the reactants and products. • Step 3: Adjust the coefficients to balance the equation.

  6. Friendly Reminder • There are a number of elements shown in the table that exist as diatomic molecules under normal conditions. When any of these elements appear in word equations, you must remember that the name refers to the diatomic molecule and insert the diatomic formula into the symbolic equation.

  7. Coefficients and Subscripts • There are two types of numbers that appear in chemical equations. There are subscripts, which are part of the chemical formulas of the reactants and products, and there are coefficients that are placed in front of the formulas to indicate how many molecules of that substance are used or produced.

  8. Example 1 • Write a balanced equation for the reaction that occurs between chlorine gas and aqueous sodium bromide to produce liquid bromine and aqueous sodium chloride. • Step 1: Write the word equation (keeping in mind that chlorine and bromine refer to the diatomic molecules). • chlorine + sodium bromide yields bromine + sodium chloride • Step 2: Substitute the correct formulas into the equation. • Cl2+NaBr  Br2+NaCl • Step 3: Insert coefficients where necessary to balance the equation. • Cl2+2 NaBr Br2+ 2 NaCl

  9. Example 2 • Write a balanced equation for the reaction between aluminum sulfate and calcium bromide to produce aluminum bromide and calcium sulfate. • Step 1: Write the word equation. • aluminum sulfate + calcium bromide yields aluminum bromide + calcium sulfate • Step 2: Replace the names of the substances in the word equation with formulas. • Al2(SO4)3+CaBr2  AlBr3+CaSO4 • Step 3: Insert coefficients to balance the equation. • products. • Al2(SO4)3 + CaBr2  2 AlBr3+CaSO4 • Al2(SO4)3 + CaBr2  2 AlBr3+3 CaSO4 • Al2(SO4)3 + 3 CaBr2  2 AlBr3+3 CaSO4 • 2 Al2(SO4)3 + 6 CaBr2  4 AlBr3 + 6 CaSO4

  10. Practice Problem 1 • Given the following skeletal (un-balanced) equations, balance them. • CaCO3(s) CaO(s)+CO2(g) • CaCO3(s) CaO(s) + CO2(g) • H2SO4(aq) + Al(OH)3(aq) Al2(SO4)3(aq) + H2O(l) • 3 H2SO4(aq) + 2 Al(OH)3(aq)  Al2(SO4)3(aq) + 6 H2O(l) • Ba(NO3)2(aq) +Na2CO3(aq)  BaCO3(aq) +NaNO3(aq) • Ba(NO3)2(aq) + Na2CO3(aq)  BaCO3(aq) +2 NaNO3(aq)

  11. Conversion of Mass in Chemical Reactions • We already know from the law of conservation of mass that mass is conserved in chemical reactions. But what does this really mean? Consider the following reaction. • Fe(NO3)3 + 3 NaOH Fe(OH)3 + 3 NaNO3 • Verify to yourself that this equation is balanced by counting the number of each type of atom on each side of the equation. We can demonstrate that mass is conserved by determining the total mass on both sides of the equation.

  12. Mass of Reactant Side • Fe(NO3)3 + 3 NaOH Fe(OH)3 + 3 NaNO3 • Mass of the Reactant Side: • 1 molecule of Fe(NO3)3 molecular weight = (1)*(241.9 daltons) = 241.9 daltons • 3 molecules of NaOHmolecular weight = (3)*(40.0 daltons) = 120. daltons • Total mass of reactants = 241.9 daltons + 120. daltons = 361.9 daltons

  13. Product Side Mass • Fe(NO3)3 + 3 NaOH Fe(OH)3 + 3 NaNO3 • 1 molecule of Fe(OH)3molecular weight = (1)*(106.9 daltons) = 106.9 daltons • 3 molecules of NaNO3molecular weight = (3)*(85.0 daltons) = 255 daltons • Total mass of products = 106.9 daltons + 255 daltons = 361.9 daltons

  14. Lesson Summary • Chemical equations must always be balanced. • Balanced chemical equations have the same number and type of each atom on both sides of the equation. • The coefficients in a balanced equation must be the simplest whole number ratio. • Mass is always conserved in chemical reactions.

  15. Practice Problem 2 • Explain in your own words why coefficients can change but subscripts must remain constant. • Which set of coefficients will properly balance the following equation: C2H6 + O2  CO2 + H2O • 1, 1, 1, 1 • 1, 3, 2, 2 • 1, 3.5, 2, 3 • 2, 7, 4, 6

  16. Practice Problem 3 • When properly balanced, what is the sum of all the coefficients in the following chemical equation: SF4 + • H2O  H2SO3 + HF • 4 • 7 • 9 • None of the above

  17. Practice Problem 4 • When the following equation is balanced, what is the coefficient found in front of the O2: P4+O2+H2O  H3PO4 • 1 • 3 • 5 • 7

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