Theoretical and Percent Yield How To Balance Chemical Equations Balancing the reaction's chemical equation is the first step in calculating theoretical and percent yield. The principle of balancing an equation is to obey the law of conservation of mass,which states simply that matter cannot be created nor destroyed, but only transferred to a different format.   When a chemical equation is balanced, it displays the ratios in which the reactants react.  In essence, the amount of atoms on one side must equal the amounts of atoms on the other.  That is, the amount of carbons on the reactant side must equal the number of carbon atoms on the product side, etc.   The first step in balancing any equation, is to write out the correct chemical formula:   For Example:           CH4 + O2 -->  CO2 + H2O Is the reaction for the combustion of methane (CH4) in excess oxygen (O2) To balance the equation, you need to find the smallest whole number coefficients so that each element is balalnced in the reaction. To solve for these coefficients, use a sytem of equations: For the current example: aCH4 + bO2 -->  cCO2 + dH2O Multiplying each element by the coefficient gives you the following equations: Eq'n 1: a = c When the equation is balanced, there will be (1 x a) Carbon atoms from methane on the reactants side and (1 x c) Carbon atoms from carbon dioxide on the products side Eq'n 2: 4a = 2d When the equation is balanced, there will be (4 x a) Hydrogen atoms from methane on the reactants side and (2 x d) hydrogen atoms from water on the products side Eq'n 2(i): 2a = d Note that Eq'n 2 can be simplified to this format Eq'n 3: 2b = 2c + d When the equation is balalnced, there will be (2 x b) oxygen atoms from oxygen gas on the reactant side, (2 x c) oxygen atoms from carbon dioxide on the products side and (1 x d) oxygen atoms from water on the product side. Eq'n 3(i): b = 2a Note that this formula results by subsitituting Eq'n 1 and Eq'n 2(i) into Eq'n 3 and simplifying To solve for a, b, c and d: First take the smallest whole number that satisfies Eq'n 1, which is simply a = c =1 Then, substitute that value into Eq'n 2(i): 2a = d 2(1) = d d = 2 and Eqn' 3(i): b = 2a b = 2(1) b = 2 Your coefficients are a = 1, b = 2, c = 1 and d = 2 When writing your balanced equation, the coefficient 1 is assumed and can be omitted, yielding the formula: CH4 + 2O2 -->  CO2 + 2H2O Now, verify your solution: Carbon: # of reactant moles: 1 # of product moles: 1 Balanced! Hydrogen: # of reactant moles: 4 # of product moles: 4 Balanced! Oxygen: # of reactant moles: 4 # of product moles: 4 Balanced! As all the element are balanced, the solution is correct.
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