RULE #1: Start with a SINGLE ELEMENT in the LARGEST COMPOUND

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1 RULE #1: Start with a SINGLE ELEMENT in the LARGEST COMPOUND Identify the compound with the MOST ATOMS. Choose a single element within the compound that is NOT hydrogen or oxygen. Hydrogen and oxygen should be balanced last when possible. Balance the single element on both sides of the equation by adjusting COEFFICIENTS. After one element is balanced, balance the rest of the equation by changing coefficients as necessary. S 8 + O 2 SO 2 Start with sulfur. There are 8 sulfur atoms on the reactant side of the equation. Place a coefficient of 8 in front of SO 2 to balance the sulfur atoms. S 8 + O 2 8 SO 2 There are now 16 atoms of oxygen on the product side of the equation. What number should be placed in front of the O 2 to reach 16 oxygen atoms on the reactant side of the equation? S O 2 8 SO 2

2 RULE #2: The Polyatomic Ion Trick Identify any POLYATOMIC IONS (Reference Table F) that appear on both sides of the equation. Use COEFFICIENTS to balance the POLYATOMIC IONS as a single unit. Balance the rest of the equation by changing coefficients as necessary. AgI + Fe 2 (CO 3 ) 3 FeI 3 + Ag 2 CO 3 There are 3 carbonate (-CO -2 ) polyatomic ions on the reactant side of the equation. Place a coefficient of 3 in front of Ag 2 CO 3 to balance the carbonates. AgI + Fe 2 (CO 3 ) 3 FeI Ag 2 CO 3 Start with (-CO 3-2 ). There are now 6 atoms of Ag on the product side of the equation. What number should be placed in front of AgI to reach 6 Ag atoms on the reactant side of the equation? 6 AgI + Fe 2 (CO 3 ) 3 FeI Ag 2 CO 3 There are now 6 iodine atoms on the reactant side of the equation. What coefficient should be place in front of FeI 3 to reach 6 iodine atoms on the product side of the equation? 6 AgI + Fe 2 (CO 3 ) 3 2 FeI Ag 2 CO 3

3 RULE #3: The Water Trick Determine whether BOTH of the following criteria are met by the unbalanced equation: o Does H 2 O appear on one side of the equation? o Does the hydroxide polyatomic ion (-OH - ) appear on the other side of the equation? Change the H 2 O to H(OH) and balance the hydrogen atom (H) and hydroxide polyatomic (-OH - ) ion as two separate entities. REMEMBER: Both criteria must be met to use this trick, so you cannot use The Water Trick for combustion reactions! H(OH) NaOH + H 2 (CO 3 ) Na 2 CO 3 + H 2 O Change H 2 O to H(OH) There are 2 hydrogen atoms on the reactant side of the equation. Place a coefficient of 2 in front of H(OH) to balance the hydrogen atoms. NaOH + H 2 (CO 3 ) Na 2 CO H(OH) There are now 2 (-OH - ) ions on the product side of the equation. What number should be placed in front of NaOH to reach 2 (-OH - ) ions on the reactant side of the equation? 2 NaOH + H 2 (CO 3 ) Na 2 CO H(OH)

4 RULE #4: The 0.5 Rule Are you working with a combustion reaction? If you start with oxygen, you re going to have a bad time. Remember: Elements that appear in multiple places on one side of an equation should be balanced last. Balance carbon. Balance hydrogen. Balance oxygen, using non-whole number coefficients (i.e., 1.5 ) if necessary. Multiply the entire equation by a number that will make all coefficients into whole numbers. C 2 H 6 + O 2 CO 2 + H 2 O Start with carbon. There are 2 carbon atoms on the reactant side of the equation. What coefficient must be placed in front of CO 2 to balance the carbon atoms? C 2 H 6 + O 2 2 CO 2 + H 2 O Balance hydrogen next. There are 6 hydrogen atoms on the reactant side of the equation. What coefficient must be placed in front of H 2 O to balance the hydrogen atoms? C 2 H 6 + O 2 2 CO H 2 O Oxygen is balanced last. There are 7 oxygen atoms on the product side of the equation. What non-whole number coefficient can be placed in front of O 2 to reach 7 oxygen atoms on the reactant side of the equation? C 2 H O 2 2 CO H 2 O Multiply all coefficients by a number that will make 3.5 into a whole number. Balanced! (C 2 H O 2 2 CO H 2 O) x 2 = 2 C 2 H O 2 4 CO H 2 O

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