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Balancing Symbolic Equations/Reactions simple inspection PRELIMINARY UNDERSTANDINGS Before an equation is balanced, the mathematical meaning of formulas must be understood. For example, NaCl represents 1 atom of sodium and 1 atom of chlorine bonded together in a particle called a molecule. In AlCl3, there is 1 atom of aluminum bonded with 3 atoms of chlorine in a particle called a molecule. Once atoms are locked together in a molecule, that ratio is not able to be changed without another chemical reaction. In more complicated molecules this ratio still holds true. For example the formula (NH4)2CO3 contains 2 ion particles of NH4 and 1 ion particle of CO3. In NH4, there is 1 atom of N and 4 atoms of H bonded together, and in CO3, there is 1 atom of C and 3 atoms of O bonded together. Again, once the 2 ion particles of NH 4 and 1 ion particle of CO3. are bonded, the ratio is fixed and cannot change unless another reaction takes place. Overall there are a total of 2 N atoms, 8 H atoms, 1 C atom and 3 O atoms. In order to indicate more than one molecule, no super or subscripts are used. Instead, full sized numbers are used and placed immediately to the left side of the molecule symbol. These numbers are called COEFFICIENTS. For example, to symbolize 4 molecules of NaCl, the correct written expression would be 4 NaCl. This means that there are a total of 4 Na atoms and 4 Cl atoms, although they are still rigidly bound in NaCl pairs. To symbolize 5 molecules of AlCl3, the correct expression would be 5 AlCl3 This means that there are a total of 5 Al atoms and 15 Cl atoms, although they are still rigidly bound in AlCl3 groups. Finally, to symbolize 3 molecules of (NH4)2CO3, the correct written expression would be 3 (NH4)2CO3. This means that there are a total of 3 (NH4)2 ion particle groups or 6 NH4 ion particle groups, and 3 CO3 ion particle groups. This means that there are a total of 6 N atoms, 24 H atoms, 3 C atoms, and 9 O atoms, although they are still rigidly bound into molecule groups of (NH4)2CO3. go to the next page Example A 1. First look at the equation: Na2CO3 + CaCl2 → CaCO3 + NaCl The total number of symbols for each element or ion depicted on the left side of the arrow MUST be equal to the total number of symbols for each element or ion depicted on the right side of the arrow. → Na2CO3 + CaCl2 CaCO3 + NaCl left side right side >two Na >one Na one CO3 one CO3 one Ca one Ca >two Cl >one Cl The small wedges mark the parts of the equation that are NOT balanced. The number of symbols of Na and Cl are NOT the same on both sides of the arrow. There not enough symbols for Na or Cl on the right side. 2. To balance this equation, coefficients are placed in the equation immediately to the left of the formula for the molecule containing the symbols needing to be balanced. The coefficient has the effect of multiplying all elements or ions in the formula by the amount of the coefficient. → Na2CO3 + CaCl2 CaCO3 + 2 NaCl left side right side >two Na one CO3 one Ca 2 x >one Na = two Na >two Cl 2 x >one Cl = two Cl one CO3 one Ca This equation is now balanced. All components have the same amount on both sides of the arrow. Final answer: Na2CO3 + CaCl2 → CaCO3 + 2 NaCl go to the next page Example B 1. First look at the equation: Na2CO3 + AlCl3 → Al2(CO3)3 + NaCl The total number of symbols for each element or ion MUST be the same on both sides of the arrow. → Na2CO3 + AlCl3 Al2(CO3)3 + NaCl left side right side >two Na >one Na >one CO3 >three CO3 >one Al >two Al >three Cl >one Cl Notice in this equation, that NOTHING is balanced. 2. To balance this equation, pick a place to start, such as Al. Place a coefficient in front of the entire molecule where Al is deficient. Although the Al is now balanced on both sides, notice that the coefficient changes the Cl total. Na2CO3 + 2x 2x → 2 AlCl3 Al2(CO3)3 + NaCl left side >two Na >one CO3 right side >one Na >three CO3 >one Al = two Al >two Al >three Cl = six Cl >one Cl 3. Since balancing Al changed the Cl as well, add a coefficient on the right side to balance the Cl. Again notice the effect this new coefficient has on Na. Na2CO3 + 2 AlCl3 → Al2(CO3)3 + left side NaCl right side 6 x >one Na = six Na >two Na >one CO3 2x 2x 6 >three CO3 >one Al = two Al >two Al 6 x >one Cl = six Cl >three Cl = six Cl go to the next page 4. Since adding the coefficient of 6 to NaCl changed the Na as well as the Cl, go to the left side and add the coefficient of 3 that will bring the Na back into balance. Again notice the effect that this new coefficient has on the CO3. 3 Na2CO3 + 2 AlCl3 left side 3x 3x 2x >two Na = six Na >one CO3 = three CO3 >one Al = two Al → Al2(CO3)3 + 6 NaCl right side 6 x >one Na = six Na >three CO3 >two Al 2x >three Cl = six Cl 6 x >one Cl = six Cl 5. The last coefficient has completed the process. All amounts of element symbols/ions are now equal. This equation is balanced. Final answer: 3 Na2CO3 2 AlCl3 + → 6 Al2(CO3)3 + NaCl WARNING ! ! ! **Coefficients are NEVER placed in the middle of a formula. They are always and only placed on the left side of a formula, and they apply ONLY to the whole formula immediately to the right of the coefficient. **When finished balancing an equation, make sure that the coefficients are the smallest possible set of coefficients that will balance the equation. ie: Na2CO3 + 2 Na2CO3 + CaCl2 2 CaCl2 → CaCO3 + → 2 CaCO 3 + 2 NaCl 4 NaCl Both equations are balanced, but the equation with the smaller set of coefficients is correct. Final correct answer is : Na2CO3 + CaCl2 → CaCO3 + 2 NaCl