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Balancing Equations

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Law of Conservation of Mass The Law of Conservation of Mass states: that mass is neither created nor destroyed in any chemical reaction. Therefore balancing equations requires the same number of atoms on both sides of a chemical reaction. The number of atoms in the Reactants must equal the Number of atoms in the Products

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Balance elements in the following order: 1. Nonmetals 2. Metals 3. Polyatomic ions 4. Hydrogen 5. Oxygen

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Polyatomic Ions When balancing polyatomic ions treat them as one unit if they appear on both sides of the equation.

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Balancing Equations Complete an atom inventory. N 2 + H 2 NH 3 __N__ __H__

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Balancing Equations Nothing is balanced. Balance the nitrogen first by placing a coefficient of 2 in front of the NH 3. N 2 + H 2 2NH 3

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Balancing Equations Hydrogen is not balanced. Place a 3 in front of H 2. N 2 + 3H 2 2NH 3 Complete a final atom inventory. __N__ __H__ 2 6

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Balancing Equations Complete an atom inventory. C 3 H 8 + O 2 → CO 2 + H 2 O __C__ __H__ __O__

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Balancing Equations Nothing is balanced. Balance the carbon first by placing a coefficient of 3 in front of the CO 2. C 3 H 8 + O 2 → 3CO 2 + H 2 O

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Balancing Equations Now balance the hydrogen by placing a 4 in front of the water on the products side. C 3 H 8 + O 2 → 3CO 2 + 4H 2 O

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Balancing Equations Balance the oxygen on the reactants side with a coefficient of 5. C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O Complete a final atom inventory. __C__ __H__ __O__

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Balancing Equations Complete an atom inventory. → BF 3 + Li 2 SO 3 → B 2 (SO 3 ) 3 + LiF __B__ __F__ __Li__ __SO 3 __

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Balancing Equations Nothing is balanced. Balance the fluorine first by placing a coefficient of 3 in front of the LiF. → BF 3 + Li 2 SO 3 → B 2 (SO 3 ) 3 + 3LiF

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Balancing Equations Now balance the boron by placing a 2 in front of the BF 3 on the reactants side. → 2BF 3 + Li 2 SO 3 → B 2 (SO 3 ) 3 + 3LiF Notice we’ve unbalanced the fluorine, and now must adjust our coefficients.

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Balancing Equations Now balance the boron by placing a 2 in front of the BF 3 on the reactants side. → 2BF 3 + Li 2 SO 3 → B 2 (SO 3 ) 3 + 6LiF Notice we’ve unbalanced the fluorine, and now must adjust our coefficients.

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Balancing Equations Balance the lithium sulfite on the reactants side with a coefficient of 3. Complete a final atom inventory. → 2BF 3 + 3Li 2 SO 3 → B 2 (SO 3 ) 3 + 6LiF __B__ __F__ __Li__ __SO 3 __

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YOU ARE BALANCED!!

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