Balancing Chemical Equations


 Franklin Parrish
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1 Balancing Chemical Equations A mathematical equation is simply a sentence that states that two expressions are equal. One or both of the expressions will contain a variable whose value must be determined by solving the equation. Linear equations are equations in which the variable being solved for is raised to a power no higher than 1. Equations in which the variable being solved for is raised to a power of 2 are called quadratic equations. The problems in this text do not require you to solve quadratic equations, so this topic will not be covered here. Consider the following linear equation. 5x  3 = 2x + 9 To solve the equation for x, the first step is to get all of the terms involving x on one side of the equation, and all other terms on the other side. We are allowed to perform the following operations on any equation without changing the equality that the equation represents. 1. Adding the same quantity to both sides of the equation. 2. Subtracting the same quantity from both sides of the equation. 3. Multiplying both sides of the equation by the same quantity. 4. Dividing both sides of the equation by the same quantity. The basic rule is: Whatever you do to one side, do to the other side also. In the equation 5x  3 = 2x + 9, we will collect all of the x s on the left side of the equation and all of the numbers on the right side. We must add 3 to each side to get the numbers on the right side, and subtract 2x from each side to get the x on the left side. Original equation: 5x  3 = 2x + 9 (1) add three to both side: x = 2x + 12 (2) subtract 2x from both sides: 2x 2x 3x = 12 Now to solve for x, we must divide both sides of the equation by 3. 1
2 (3) divide both sides by 3: 3x = Final Answer: x = 4 Balancing Equations Example 1 Solve for the value of x in the equation below. Solution: 6x + 7 = 4x + 23 First, subtract 7 from both sides. Then, subtract 4x from both sides. Finally, divide both sides by 2 to get the final answer: x = 8. You will be on track to get the right answer if you remember one simple rule: Whatever you do to one side, do to the other side also. Now what does this have to do with balancing equations? You can use basic algebra rearrangements like we did in the example above to help you balance equations. Look at this equation. Ni(s) + HCl(aq) NiCl 2 (aq) + H 2 (g) A balanced equation must have the same number of each kind of atom on both sides of the equation. The same number of nickel atoms must appear on both sides of the equation. The same number of hydrogen atoms must appear on both sides of the equation. The same number of chlorine atoms must appear on both sides of the equation. As it is currently written, the equation is not balanced. The nickel atoms are balanced, but the hydrogen and chlorine atoms are not balanced. Let s balance the chlorine atoms using some algebra. There is one chlorine atom on the left side of the equation and two chlorine atoms on the right side of the equation. What do you need to multiply the left side of the equation by in order to make the two sides of the equation have the same number of chlorine atoms? Let this unknown multiplication factor be x. We can set up an algebraic equation for this question that looks like: x(1) = 2 2
3 To solve for x, divide both sides by one. The final answer is x = 2. If we put a coefficient of two in front of HCl, the equation becomes: Ni(s) + 2HCl(aq) NiCl 2 (aq) + H 2 (g) The equation is now balanced. That was a very simple application of algebra to the problem of balancing an equation. Let s look at a more complex example. Fe(s) + O 2 (g) Fe 2 O 3 (s) Let s balanced the iron atoms first. There is one iron atom on the left side of the equation and two iron atoms on the right side of the equation. The algebra expression that represents this situation would be: x(1) = 2 Solving for x by dividing both sides by one gives us the final answer of: x = 2. We need to put a coefficient of two in front of Fe(s) on the left to balance the iron atoms. 2 Fe(s) + O 2 (g) Fe 2 O 3 (s) There are two oxygen atoms on the left side of the equation and three oxygen atoms on the right side of the equation. The algebra expression that represents this situation would need to say that some unknown multiplication factor, y, times the two oxygen atoms on the left will equal the three oxygen atoms on the right. y(2) = 3 Solve for y by dividing both sides by two. The final answer is y = 3/2. We need to put a coefficient of 3/2 in front of the O 2 on the left to balance the oxygen atoms. It is easier to leave this answer in the fractional form rather than convert it to a decimal number. You ll see why soon. 2 Fe(s) + 3/2 O 2 (g) Fe 2 O 3 (s) The equation is now balanced. However, as often happens, a fraction appears in the balanced equation. Chemical reactions, except in a few limited circumstances, should not be balanced with fractions. In order to eliminate the fraction, we need to multiply all of the coefficients by the number in the denominator. Remember the basic rule: Whatever you do to one side, do to the other side also. 2. 2Fe(s) /2 O 2 (g) 2 Fe 2 O 3 (s) 3
4 The equation becomes: 4 Fe(s) + 3 O 2 (g) 2 Fe 2 O 3 (s) Balancing Equations Example 2 Balance this equation: HF(g) + SiO 2 (s) SiF 4 (g) + H 2 O(l) Solution: If chemical equation is difficult to balance, you can use simple algebra expressions to make balancing the equation easier. To balance the fluorine atoms, you could use the algebra expression: x(1) = 4 Solving for x gives you a final answer of: x = 4. You need a coefficient of four in front of HF(g). To balance the oxygens, you could use the algebra expression: 2 = y(1) Solving for y gives you a final answer of: 2 = y. You need a coefficient of two in front of H 2 O(l). The balanced equation is: 4HF(g) + SiO 2 (s) SiF 4 (g) + 2H 2 O(l) 4
5 Balancing Equations Example 3 Balance the equation C 2 H 6 (g) + O 2 (g) CO 2 (g) + H 2 O(l) Solution: Begin by balancing the carbons. The algebra expression for carbons could be 2 = x(1). Solving for x gives the final answer: 2 = x. You need a coefficient of two in front of CO 2 (g). The algebra expression for hydrogen could be 6 = y(2). Solving for y gives the final answer: 3 = y. You need a coefficient of 3 in front of H 2 O(l). C 2 H 6 (g) + O 2 (g) 2 CO 2 (g) + 3 H 2 O(l) The algebra expression for oxygen is more complex. There are two oxygen atoms on the left side of the equation and oxygen atoms on the right side of the equation. The algebra expression could be: z(2) = 7 Solving for z gives the final answer: z = 7/2. The balanced equation is: C 2 H 6 (g) + 7/2 O 2 (g) 2 CO 2 (g) + 3 H 2 O(l) Although the equation is balanced, we need to eliminate the fraction from the chemical equation. We do this by multiplying coefficients on both sides of the equation by whatever value is in the denominator. In this case, we would multiply by two. 2 C 2 H 6 (g) /2 O 2 (g) 2. 2 CO 2 (g) H 2 O(l) The equation becomes 2 C 2 H 6 (g) + 7 O 2 (g) 4 CO 2 (g) + 6 H 2 O(l) 5
6 Hints for Balancing Equations: Do I need to use algebra to balance an equation? No. Most students quickly learn how to balance the simpler chemical equations. However, if a equation is difficult or tricky to balance, writing out algebra expression for each different kind of atom is a reliable way to balance an equation. 6
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