Question

What Is Reaction Of 5C3H8+2I2+HIO3→???? ​

Answer 1

Explanation:

This is a so-called comproportionation reaction.

Explanation:

Iodic acid is reduced to elemental iodine:

HI+VO3+5H++5e−→12I2+3H2O (i)

Iodide is oxidized to elemental iodine:

I−→12I2+e− (ii)

So (i)+5×(ii)=

HI+VO3+5H++5I−→3I2+3H2O

This is AFAIK balanced with respect to mass and charge, and therefore a reasonable representation of chemical reality.

The balanced equation is

HIO3+5HI→3I2+3H2O

Explanation:

We start with the unbalanced equation:

HIO3+HI→I2+H2O

Step 1. Identify the atoms that change oxidation number

We start by determining the oxidation numbers of every atom in the equation.

+1H+5I-2O3++1H-1I→0I2++1H2-2O

+1H+5I-6O3++1H-1I→0I2++2H2-2O

We see that the oxidation number of I in HIO3 is reduced to 0 in I2 and the oxidation number if I in HI is increased to 0 in I2.

This is a comproportionation reaction, a reaction in which an element in a higher oxidation state reacts with the same element in a lower oxidation state to give the element in an intermediate oxidation state.