That's because I knew that there are 13 hydrogens in the C 5H 11NH 2 and that meant a 13⁄ 2 in front of the water. Notice that I used a 2 in front of C 5H 11NH 2. The chemical equation is balanced in a chemically-correct sense with the fractional coefficients. The final step towards whole number coefficients is just a convention. We can certainly have 3⁄ 2 of a mole of carbon atoms or 3⁄ 2 of a mole of carbon dioxide molecules. The chemical reality of atoms reacting in ratios of small whole numbers is reflected in the final answer.Īnother way to look at the coefficients is in terms of moles.
What's that you say? You can't have 3⁄ 2 of an atom? Ah, just you wait.Ĥ) Multiply through by two for the final answer:ĢFe 2O 3(s) + 3C(s) -> 4Fe(s) + 3CO 2(g)Ĭomment: one way to look at this is that using the 3⁄ 2 was just a mathematical artifice to balance the equation.
Note the 3⁄ 2 in front of the C and the CO 2. With this last step, the oxygen is also balanced and the Mn was never mentioned because it started out balanced and stayed that way. Note how the hydrogen started out balanced, but the balancing of oxygen affected the hydrogen, which we addressed in the second step. Note that the vanadium was not addressed because it stayed in balance the entire time. There are a total of 5 on the right-hand side, so we put 5 on the left:Ģ) Clear the fraction by multiplying through by 2: Problem #4: FeS 2 + Cl 2 -> FeCl 3 + S 2Cl 2ġ) See how the Fe and the S are already balanced? So, look just at the Cl. It shows up a lot in balancing problems (if you haven't already figured that out!).Ĥ) Oops, that messed up the lithium, so we fix it:ġ) Balance the oxygen with a fractional coefficient (Zn and S are already balanced):Ģ) Multiply through to clear the fraction: See how the H comes only in groups of 3 on the left and only in groups of 2 on the right? Do this: Problem #2: Li + H 3PO 4 -> H 2 + Li 3PO 4Ģ) Now, look at the hydrogens.
The other element (Mg or O, depending on which one you picked) also gets balanced in this step. Problem #1: FeCl 3 + MgO -> Fe 2O 3 + MgCl 2ġ) Balance the Cl (note that 2 x 3 = 3 x 2):Ģ) Pick either the O or the Mg to balance next: Problems #1 - 10 Twenty examples Problems 11-25 Problems 26-45 Problems 46-65 Six "balancing by groups" problems Only the problems Return to Equations Menu Sixteen balance redox equations by sight ChemTeam: Balancing Chemical Equations: Problems #1 - 10
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