K2MnF6 + SbF5 ⟶ KSbF6 + MnF3 + F2What is the sum of the coefficients of the products after balancing?

To find the sum of the coefficients of the products after balancing the equation, we will follow these steps:

1. Break Down the Problem

We have the unbalanced chemical equation: $\text{K}_2\text{MnF}_6 + \text{SbF}_5 \rightarrow \text{KSbF}_6 + \text{MnF}_3 + \text{F}_2$

2. Relevant Concepts

To balance a chemical equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation.

3. Analysis and Detail

Let's count the number of each type of atom on both sides initially:

Reactants:

• K: 2
• Mn: 1
• F: 6 (from K2MnF6) + 5 (from SbF5) = 11
• Sb: 1

Products:

• K: 1 (from KSbF6)
• Sb: 1
• Mn: 1
• F: 6 (from KSbF6) + 3 (from MnF3) + 2 (from F2) = 11

Balancing Steps:

1. Balance Potassium (K): We have 2 K on the reactants side and 1 K on the products side. We can place a coefficient of 2 in front of KSbF6. $\text{K}_2\text{MnF}_6 + \text{SbF}_5 \rightarrow 2\text{KSbF}_6 + \text{MnF}_3 + \text{F}_2$

2. Recount the atoms: Products now become:

   • K: 2 (from 2 KSbF6)
   • Sb: 2 (from 2 KSbF6)
   • Mn: 1
   • F: 12 (6 from 2 KSbF6, 3 from MnF3, 2 from F2) = 11

3. Adjust Antimony (Sb): We still have too many Sb in the products, so we adjust by placing a coefficient of 2 in front of SbF5 on the reactant side: $\text{K}_2\text{MnF}_6 + 2\text{SbF}_5 \rightarrow 2\text{KSbF}_6 + \text{MnF}_3 + \text{F}_2$

4. Final Check: After balancing:

   • Reactants: K=2, Mn=1, F=16, Sb=2
   • Products: K=2, Sb=2, Mn=1, F=16

The equation is now balanced.

4. Verify and Summarize

Sum of the coefficients of the products:

• For $2 \text{KSbF}_6$: 2
• For $\text{MnF}_3$: 1
• For $\text{F}_2$: 1

Sum = $2 + 1 + 1 = 4$

Final Answer

The sum of the coefficients of the products after balancing is $4$.