Is this equation correct?
KI + HCl +KMnO4 --> KCl +MnCl +H2O + I2
If so what would be the color of the solution?
PS. I am aware the equation is not balanced. =]
Nor is it correct.
MnCl2 is manganese(II) chloride.
Otherwise it looks ok.
KMnO4, as you no doubt know, is purple. MnCl2 is a VERY light pink color but it takes more than just a little to color a solution.
Thanks for your help.
To determine whether the equation is balanced, we need to count the number of atoms on both sides of the equation. Let's start by balancing the elements individually.
On the reactant side:
- We have one K on the left side and one K on the right side.
- We have one I on the left side and one I2 on the right side.
On the product side:
- We have one Mn on the left side and one Mn on the right side.
- We have one Cl on the left side and one Cl on the right side.
- We have two H on the left side and two on the right side.
- We have one O on the left side and none on the right side.
- We have four O on the right side but none on the left side.
To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides. Here's the balanced equation:
2KI + 8HCl + KMnO4 --> 2KCl + MnCl2 + 4H2O + 5I2
Now, regarding the color of the solution. When iodine (I2) is formed, it typically gives the solution a brownish color. So, in this reaction, the solution will likely turn brown due to the formation of iodine.