Calculate the number of moles of KIO3 consumed by reaction (2.1). Remember, it is the limiting reagent. Using this information, calculate the number of moles of I- that must have been consumed by reaction.
Unbalanced equation: IO -(aq) + I-(aq) + H+(aq) ÿ I (aq) + H O(l)
Concentration of KIO3 is 0.1201 mol/l
Volume of KIO3 is 5.00ml or 0.005 L
I got the amount of mols for KIO3 but have no idea how to get it for I-
Hey this is the same uoit lab I just completed and submitted, if you can tell me your value again for KLO3 (amount of mols) i'll be able to solve it for you
repost, did you do 0.1201 * 0.005=6*10^-4, if so then you'd subtract this number from the one calculated in part 2 and there's your answer
Well, let's start by balancing the equation:
IO3-(aq) + 5I-(aq) + 6H+(aq) -> 3I2(aq) + 3H2O(l)
Now that the equation is balanced, we know that for every 1 mole of KIO3, we need 5 moles of I- to react. Since KIO3 is the limiting reagent, we can calculate the number of moles of I- that must have been consumed by using the ratio:
Moles of I- = Moles of KIO3 x (5 moles of I- / 1 mole of KIO3)
Given that the concentration of KIO3 is 0.1201 mol/L and the volume is 0.005 L, we can calculate the number of moles of KIO3:
Moles of KIO3 = Concentration of KIO3 x Volume of KIO3
Moles of KIO3 = 0.1201 mol/L x 0.005 L
Now, let's plug this into the ratio:
Moles of I- = (0.1201 mol/L x 0.005 L) x (5 mol of I- / 1 mol of KIO3)
Moles of I- = 0.0003 mol x 5
Moles of I- = 0.0015 mol
Therefore, the number of moles of I- that must have been consumed by the reaction is 0.0015 mol.
To calculate the number of moles of I- consumed by the reaction, we need to determine the stoichiometry of the reaction. The unbalanced equation provided is:
IO- (aq) + I- (aq) + H+ (aq) → I2 (aq) + H2O (l)
To balance the equation, we need to add coefficients to balance the number of atoms of each element on both sides of the equation. Balancing this equation gives:
5 IO- (aq) + 6 I- (aq) + 6 H+ (aq) → 3 I2 (aq) + 3 H2O (l)
Now, we have a balanced equation with a stoichiometric ratio. According to the balanced equation, 5 moles of IO- reacts with 6 moles of I-.
Given that KIO3 is the limiting reagent, we can use its stoichiometric ratio to calculate the number of moles of I- consumed.
You mentioned that the concentration of KIO3 is 0.1201 mol/L, and the volume is 0.005 L. Therefore, the number of moles of KIO3 consumed is:
0.1201 mol/L * 0.005 L = 0.0006005 moles of KIO3
Since the stoichiometric ratio between KIO3 and I- is 5:6, we can calculate the number of moles of I- consumed as:
(6/5) * 0.0006005 moles of KIO3 = 0.0007206 moles of I-
Therefore, the number of moles of I- consumed by the reaction is approximately 0.0007206 moles.