C2H5OH + 6NaOH + 412→ ..CHI) + ..HCOONa + ...Nal + ...H20

To balance this chemical equation, we need to ensure that the number of each type of atom is the same on both sides of the equation.

Given reaction: C2H5OH + 6NaOH → XCHI + YHCOONa + ZNaI + WH2O

First, let's determine the number of each type of atom on each side:

• Carbon (C): 2 (left) = X (right)
• Hydrogen (H): 6 (left) = 5X + W (right)
• Oxygen (O): 2 (C2H5OH) + 6 (NaOH) = 2Y + 2Z + W
• Sodium (Na): 6 (NaOH) = 1Z
• Iodine (I): 0 (left) = X + Z
• Formate ion (HCOO-): 0 (left) = Y
• Water (H2O): 0 (left) = W

Now, let's simplify and balance the equation:

• Carbon (C): 2 (left) = X (right) => X = 2
• Hydrogen (H): 6 (left) = 5X + W (right) => W = 1
• Oxygen (O): 2 (C2H5OH) + 6 (NaOH) = 2Y + 2Z + W => 2 + 6 = 2Y + 2Z + 1 => 8 = 2Y + 2Z + 1 => 7 = 2Y + 2Z
• Sodium (Na): 6 (NaOH) = 1Z => Z = 6
• Iodine (I): 0 (left) = X + Z => 0 = 2 + 6 => Not balanced
• Formate ion (HCOO-): 0 (left) = Y => Y = 0

The balanced equation is: C2H5OH + 6NaOH → 2CHI + HCOONa + 6NaI + H2O