I am so confused with my chemistry lab right now. We were given the following equations to work with:

[Equation 1] 2 I^- + H2O2 + 2 H^+ ----> I2 + 2 H2O
[Equation 9] 2S2O3^2- + I2 ----> 2I^- +S4O6^2-

And During the Experiment we used the following for each reaction:
T1 | 10 mL of 0.3 M KI | 10mL of 0.02 M Na2S2O3 | 30 mL of 0.1 M H2O2
T2 | 20 mL of 0.3 M KI | 10mL of 0.02 M Na2S2O3 | 30 mL of 0.1 M H2O2
T3 | 30 mL of 0.3 M KI | 10mL of 0.02 M Na2S2O3 | 30 mL of 0.1 M H2O2
T4 | 30 mL of 0.3 M KI | 10mL of 0.02 M Na2S2O3 | 50 mL of 0.1 M H2O2
T5 | 30 mL of 0.3 M KI | 10mL of 0.02 M Na2S2O3 | 70 mL of 0.1 M H2O2

For the lab we also used water, a buffer, and starch in the solutions.

(From lab manual - Calculate the number of moles of S2O3^2- that is consumed in each reaction. Use stoichiometry in equation 9. From equation 9 one-half the moles of S2O3^2- equals the moles of I2 that form in the reaction. Record the moles of I2 formed during the elapsed time)

I have no idea where to start... So, to calculate the number of moles of S2O3^2-, would I have to take 0.02M Na2S2O3 and multiply it by the .010L (10 mL) that we used for the solutions to find the moles - which would be 2 x 10^-4 mol Na2S2O3. But how would I find the amount consumed. Would all of it be consumed in the end? Because in this lab, once all the S2O3^2- is consumed the solution turns dark blue. And how do I find I2. Some help would be really appreciated.

All of the numbers run together with no spaces so it's hard to know what you've done. However, equation 1 is a reaction that FORMS I2. To know how much is formed you titrate it with standard S2O3^2-. Yes, mols S2O3^2- = M x L and mols I2 which came from equation 1 will be 1/2 the thiosulfate mols. You titrate all of the I2 with thiosulfate and that's when the starch turns blue. Again, that's the amount of I2 formed from the first reaction.

To calculate the number of moles of S2O3^2- consumed in each reaction, you can use the stoichiometry of Equation 9. According to the balanced equation:

2S2O3^2- + I2 ----> 2I^- +S4O6^2-

It shows that 2 moles of S2O3^2- reacts with 1 mole of I2. Therefore, one-half of the moles of S2O3^2- that react is equal to the moles of I2 formed in the reaction.

To calculate the moles of S2O3^2- consumed:

1) You correctly calculated the initial moles of Na2S2O3 using the molarity and volume: 0.02 M x 0.01 L = 0.0002 mol Na2S2O3.

2) Since the stoichiometry states that 2 moles of S2O3^2- react with 1 mole of I2, you divide the initial moles of Na2S2O3 by 2: 0.0002 mol Na2S2O3 / 2 = 0.0001 mol S2O3^2-. This represents the moles of S2O3^2- consumed in the reaction.

In regards to the consumption of S2O3^2-, once all of it is consumed, the solution turns dark blue. This color change indicates the endpoint of the reaction, and no more S2O3^2- is present.

Now, to find the moles of I2 formed during the reaction, you need to determine the moles of S2O3^2- consumed using the above calculation. Since the stoichiometry states that 1 mole of I2 is produced for every 2 moles of S2O3^2- consumed, you can conclude that the moles of I2 formed are also 0.0001 mol.

Remember to repeat these calculations for each reaction (T1, T2, T3, T4, T5) and record the respective moles of I2 formed during each elapsed time.

I hope this clarifies the steps and assists you in your chemistry lab.

To calculate the number of moles of S2O3^2- consumed in each reaction, you need to use stoichiometry based on Equation 9. From Equation 9, we know that 2 moles of S2O3^2- reacts with 1 mole of I2.

First, you correctly calculated the number of moles of Na2S2O3 in the solution. For each trial, you have 0.02 M Na2S2O3 and 10 mL of it. To find moles of Na2S2O3, you can multiply the concentration (0.02 M) by the volume in liters. So, 0.02 M x 0.010 L = 2 x 10^-4 mol Na2S2O3.

Since the stoichiometry tells us that 2 moles of S2O3^2- reacts with 1 mole of I2, we can conclude that for every 2 x 10^-4 mol of Na2S2O3 consumed, we will form 1 x 10^-4 mol of I2.

To find the amount of I2 formed during the elapsed time, you need to multiply the moles of I2 formed by the molar mass of I2. The molar mass of I2 is approximately 253.8 g/mol. Therefore, to calculate the mass of I2 formed, you multiply the moles of I2 by the molar mass: 1 x 10^-4 mol I2 x 253.8 g/mol = 0.02538 g I2.

Keep in mind that these calculations assume the reaction is complete, meaning all the S2O3^2- is consumed. However, if the solution turns dark blue before you add all of the Na2S2O3, it suggests that the reaction has reached its endpoint and all the S2O3^2- has been consumed.

I hope this helps! Let me know if you have any further questions.