# The reaction below takes place until the solution went from yellow to clear. If iodine was the...

## Question:

The reaction below takes place until the solution went from yellow to clear. If iodine was the contributor to the yellow color, how many moles of I{eq}_2 {/eq}(aq) are present after completion? Explain your answer.

CH{eq}_3 {/eq}COCH{eq}_3 {/eq}(aq) + I{eq}_2 {/eq}(aq) --> CH{eq}_3 {/eq}COCH{eq}_2 {/eq}I(aq) + H{eq}^+ {/eq}(aq) + I{eq}^+ {/eq}(aq)

## Balancing Equation

A chemical needs to be balanced if quantitative calculations are to be done regarding the amount of the chemicals. Balancing a reaction by changing the coefficients before the chemical formulas so the number of each element on the reactant and product sides are equal.

zero moles of {eq}I_2 {/eq} remains after reaction completion.

If {eq}I_2 {/eq} is responsible for the yellow color and we are told that the solution when from yellow to clear (which probably suppose to be colorless), we can assume there are no {eq}I_2 {/eq} left after the completion of the reaction.