Presentation on theme: "Reactions in Chemistry Balancing Equations. Balanced equations Balanced equations reflect the fact that matter is neither created nor destroyed in a chemical."— Presentation transcript:
Balanced equations Balanced equations reflect the fact that matter is neither created nor destroyed in a chemical reaction. An equation is balanced if the same number of each type of atom is found in both the reactants and products of the reaction. A total count must be taken on each side and the numbers compared to determine whether or not an equation is balanced.
Counting reactants There are two reactants in the following reaction. Determine the number of atoms of each type that are listed on the reactant side: CaBr 2 :AgNO 3 : Ca –Ag – Br –N – O –
Counting products There are two products in this reaction. Determine the number of atoms of each type that are listed on the product side: AgBr:Ca(NO 3 ) 2 : Ag –Ca – Br –N – O –
Counting products As you can see below, in a properly balanced equation there will be the same number of each type of atom in the reactants and products: REACTANTS:PRODUCTS: CaBr 2 : AgNO 3 :AgBr:Ca(NO 3 ) 2 Ca – 1 Ag – 2Ag – 2Ca – 1 Br – 2 N – 2Br – 2N – 2 O – 6 O – 6
Counting reactants Notice that in the following reaction oxygen shows up in both of the reactants; the total count of reactant atoms, once these are combined, is: FeSO 4 :Ba(NO 3 ) 2 All reactant atoms: Fe – 1Ba – 1 Fe – Ba – S – 1N – 2 S – N – O – 4O – 6 O –
Counting reactants Remember this as you count the atoms on the product side as well: BaSO 4 :Fe(NO 3 ) 2 All product atoms: Ba – Fe – Fe – Ba – S – N – O – O – O –
Comparing the counts As we compare the counts of reactant and product atoms, we see that they are the same: All reactant atoms: All product atoms: Fe – 1 Ba – 1 Fe – 1Ba – 1 S – 1N – 2 O – 10
Balancing equations What if you are given a chemical equation showing the reactants and products and are asked to balance it? In order to balance the equation, we must insert coefficients that will make the count of atoms the same on the two sides of the reaction. We CANNOT change the formulas themselves, because then we would be describing a different reaction taking place between different chemicals.
Why can’t we change formulas? What if one of the reactants was water ( H 2 O ) and we needed another oxygen atom to balance the equation? If we changed the formula so that it had 2 atoms of oxygen, we wouldn’t have water any more! wps.prenhall.com
Balancing equations First, take an atom inventory. Pick an element to work with and see what coefficient could be inserted in front of a formula in order to make that element have the same number on both sides. Redo your inventory and see how the atoms of reactants and products compare now. Keep doing this until balanced. Make sure the final coefficient ratio is simplified.
Take an atom inventory Balance the following equation: ReactantsProductsBr – Na – I – Note that there are different amounts of bromine and iodine atoms on each side.
Pick an element to work with Let’s choose bromine. We need to get 2 bromine atoms on the product side. We CANNOT change the formula. Try putting a coefficient in front of the formula that contains the bromine atoms. How does this affect the count?
Redo your inventory ReactantsProductsBr – Na – I – Now our sodium atoms don’t match. How can we get them to be the same?
Pick another element to work with To get the sodium counts to match, we need to put a coefficient in front of the compound containing sodium on the reactant side. ReactantsProductsBr – Na – I –
Check for simplified ratio Now that all of the counts match, showing that there are the same number of each type of atom on each side (no matter is being created or destroyed), we need to check the coefficient ratio and see if it can be simplified: The ratio is 1:2:2:1. These numbers have no common factors, so the ratio is in simplified form.
Check for simplified ratio Why must we perform this step? The equation is also balanced if written as: However, notice that the coefficient ratio is 2:4:4:2, all of which are divisible by 2, so it would reduce to 1:2:2:1. We always write equations with the smallest coefficients that will balance the equation.
Try another example Now try balancing the following equation, remembering to check your full atom count each time you insert a new coefficient: ReactantsProductsAg – N – O – H – S –
Example 3 Do an initial inventory for the following equation: ReactantsProductsAl – Cl – If coefficients must be whole numbers, how do we get the number of chlorine atoms to match?
Example 3 What we need is a number that is a multiple of both 2 and 3. The least common multiple of 2 and 3 is 6. How do we get 6 chlorine atoms on each side? ReactantsProductsAl – Cl –
Example 4 Now try balancing the following equation, remembering to check your full atom count each time you insert a new coefficient: ReactantsProductsFe – O –
Challenge Problem #1 Now try balancing the following equation. (HINT: If a polyatomic ion appears unchanged on both sides of the equation, you can treat it as a single unit.) ReactantsProductsAl – SO 4 – Ca – OH –
Challenge Problem #2 Now try balancing the following equation. (HINT: Begin by balancing elements that appear only once on each side of the equation.) ReactantsProductsFe – O – H –
Challenge Problem #3 Now try balancing the following equation. ReactantsProductsC – H – O – Not all equations end up with small coefficients!
Balancing summary DO NOT change formulas, only coefficients! Redo inventory each time a coefficient is changed. It is generally safe to leave polyatomics together if they appear in both the reactants and products. Begin balancing elements that only appear once on each side of the equation. Be sure the final coefficient ratio is simplified.