Balancing the tri-iodomethane equation!

1. Balance $\ce{C}$ atoms: $\ce{CH3COR -> CHI3 + O^{-}COR}$
2. We note that $\ce{R}$ atoms are balanced.
3. Balance $\color{#D61F06}{\ce{H}}$ and $\color{#3D99F6}{\ce{O}}$ atoms: $\ce{CH3COR + } c\ce{OH- -> CHI3 + } f \ce{H2O + O^{-}COR}$ $\space \Rightarrow \color{#D61F06}{c+3 = 2f+1}$ and $\color{#3D99F6}{c+1 = f+2}$ $\space \Rightarrow f = 3$ and $c = 4$
4. Balance $\color{#D61F06}{\ce{I}}$ atoms and $\color{#3D99F6}{\ce{charges}}$ : $\ce{CH3COR + }b\ce{I2 + 4OH- -> CHI3 +} e \ce{I- + 3H2O + O^{-}COR}$ $\color{#D61F06}{2b= e+3}$ and $\color{#3D99F6}{4 = e+1}$ $\space \Rightarrow e = 3$ and $b = 3$

$\Rightarrow \ce{CH3COR + 3I2 + 4OH- -> CHI3 + 3I- + 3H2O + O^{-}COR} \\ \Rightarrow a + b + c + d + e + f = 1 + 3 + 4 + 1 + 3 + 3 + 1 = \boxed{16}$

Well obviously, one way to know the answer is by Google... But to understand the true art of balancing ANY equation, we can use a system of linear equations.

Hence, by comparing number of X atom on both sides:

Carbon : $2a = d + g$

Oxygen : $a + c = f + 2g$

Hydrogen : $3a + 4c = d + 2f$

R group : $a = g$

Iodine : $2b = 3d + e$

However, there are only 5 equations with 7 unknown. Now, note that the charges must be equal by the principle of conservation of charge.

Hence, $c = e + g$

Thus, solving this system, we can hence get this particular solution set : $a = t , b = 3t , c = 4t, d = t, e = 3t, f = 3t, g = t$ Hence, at $t = 1$ , the final answer is $\boxed{16}$