Last week I posted on the three methods to balance chemical equations. My quick google search seems to point towards Method 1 (by inspection or trial-error-method) and Method 2 (writing down atom counts) being the most popular methods when it comes to balancing equations. Most of the students I have worked with only knew Method 1 from their high school science/chemistry class. There’s nothing wrong with that, but I think it gets a little daunting when it comes to having to keep track of multiple numbers of atoms (like 5 or more) and it will prolong the process unnecessarily. I mean, who wants to spend 30 minutes trying to balance ONE equation??? Seriously, that 30 minutes is probably better-spent doing other things like watching YouTube, checking out latest posts in FB, Instagram, Twitter, uploading pictures to Snapchat, level up in that favorite game of yours or doing other things that will increase your happiness.

So, imagine if you have a real complex equation like this: K_{4}[Fe(SCN)_{6}] + K_{2}Cr_{2}O_{7} + H_{2}SO_{4} → Fe_{2}(SO_{4})_{3} + Cr_{2}(SO_{4})_{3} + CO_{2} + H_{2}O + K_{2}SO_{4} + KNO_{3}

which has 8 types of atoms in the equation: K, Fe, S, C, N, Cr, O and H. How long do you think it will take to balance it?

**I tried balancing it using Method 1.** I started with K since it’s the first atom I encounter from the left. Looks like I have 6 K altogether. Ok…what about on the right? Looks like a total 3 K on the right. First question, how should I balance K? I know I’ll need to “double” the K counts on the right, but which term should I start with – K_{2}SO_{4} or KNO_{3}? Not sure. Never mind…leave K alone for now. Let’s move on to Fe, it seems easier. I have 1 on the left and 2 on the right. Great! I can do this….add a “2” on the left, in front of K_{4}[Fe(SCN)_{6}], so now I have:

2 K_{4}[Fe(SCN)_{6}] + K_{2}Cr_{2}O_{7} + H_{2}SO_{4} → Fe_{2}(SO_{4})_{3} + Cr_{2}(SO_{4})_{3} + CO_{2} + H_{2}O + K_{2}SO_{4} + KNO_{3}.

Fe is balanced, for now. 7 more atoms to go…

S looks tough since I see it in multiple terms on both sides. Save it for later. Move on to C. I have 12 on the left. There’s only 1 on the right. That means I’ll place a “12” in front of CO_{2}.

2 K_{4}[Fe(SCN)_{6}] + K_{2}Cr_{2}O_{7} + H_{2}SO_{4} → Fe_{2}(SO_{4})_{3} + Cr_{2}(SO_{4})_{3} + 12 CO_{2} + H_{2}O + K_{2}SO_{4} + KNO_{3}

Awesome, 2 down, 6 more to go…

My next atom will be N. Piece of cake, kinda like C – 12 on the left and 1 on the right. So, add “12” in front of KNO_{3}.

2 K_{4}[Fe(SCN)_{6}] + K_{2}Cr_{2}O_{7} + H_{2}SO_{4} → Fe_{2}(SO_{4})_{3} + Cr_{2}(SO_{4})_{3} + 12 CO_{2} + H_{2}O + K_{2}SO_{4} + 12 KNO_{3}

Gaining momentum and confidence… 3 down, 5 more to go.

H will be next. 2 on the left, 2 on the right. Well, look at that! It’s already balanced! Moving on to Cr. There’s 2 on the left and 2 on the right. Well, how about that?? Is it my lucky day or what?

So far I have balanced Fe, C, N, H and Cr. That only took 1 minute so far. What’s 3 more atoms (K, S and O), right? WRONG!!! Those 3 that are left are tough and that’s because they appear in multiple terms all over the place. I mean look at K, on the left, it appears in K_{4}[Fe(SCN)_{6}] and K_{2}Cr_{2}O_{7}, and on the right, I see it in K_{2}SO_{4} and KNO_{3}. That’s considered the easier atom among the three. Look at O. On the left, it’s in K_{2}Cr_{2}O_{7} and H_{2}SO_{4}. On the right, it’s in Fe_{2}(SO_{4})_{3}, Cr_{2}(SO_{4})_{3}, CO_{2}, H_{2}O, K_{2}SO_{4 }and KNO_{3}. By the way, that’s EVERY single term on the right! I don’t know bout you, but I know I have better use of my time than to play trial and error with these 3 atoms.

It’s not going to be much better if I use **Method 2** either. I mean, it will help in terms of keeping track on the number of atoms, but I’ll still need to do perform trial and error with the three atoms.

So, I moved on to **Method 3**. Took me 3 minutes to set up the algebraic equations and few more minutes to solve them all. So, what do you think is the final answer? Here it is:

6 K_{4}[Fe(SCN)_{6}] + 97 K_{2}Cr_{2}O_{7} + 355 H_{2}SO_{4} → 3 Fe_{2}(SO_{4})_{3} + 97 Cr_{2}(SO_{4})_{3} + 36 CO_{2} + 355 H_{2}O + 91 K_{2}SO_{4} + 36 KNO_{3}

Here’s the video walking through the 4 steps in solving this complex equation:

hamidreza bolouriactually the 4th coeff must be 3

6 Fe2(SO4)3 => 3 Fe2(SO4)3

Dr. KYou’re right! Thanks for pointing out the error!