For chem we did a kinetics lab based on the rate of reaction of peroxydisulfate ion with iodide ion.
I'm supposed to find the amount of S2O3 ^2- consumed for every trial (in moles)
If someone could just explain how to do that with the information I currently have, that would be perfect!
The time taken for the consumption of N2S2O3 at 1.0ml is 2:46 (min:sec). The concentration of Na2S2O3 is 0.20M. There is 0.0002 moles of Na2S2O3 present in a 1.0ml aliquot of Na2s2O3 solution.
I may be overthinking this. Sorry.
mols = M x L
So it would be (0.20M)(1.0ml/1000L)=0.0002moles of S2O3 consumed ?
For each mL you used it is that, yes. I don't know the details of the experiment but if all of the 1 mL was consumed then the amount consumed is M x L = ?, yes.
No worries, let's break this down step by step.
To find the amount of S2O3^2- consumed for each trial, you need to use the information provided in the lab.
First, calculate the time taken for the consumption of N2S2O3 for each trial in seconds. In this case, 2:46 is equal to 2 minutes and 46 seconds, which can be converted to seconds by multiplying the number of minutes by 60 and then adding the number of seconds:
2 minutes * 60 seconds/minute = 120 seconds
120 seconds + 46 seconds = 166 seconds
So, the time taken for the consumption of N2S2O3 at 1.0 mL is 166 seconds.
Next, you need to determine the amount of Na2S2O3 used in each trial. You are given that the concentration of Na2S2O3 is 0.20 M (moles per liter) and there is 0.0002 moles of Na2S2O3 present in a 1.0 mL aliquot of Na2S2O3 solution.
To find the moles of Na2S2O3 consumed, you can use the formula:
moles of Na2S2O3 = concentration of Na2S2O3 * volume of Na2S2O3 (in liters)
Since the volume is given in milliliters, you need to convert it to liters. One milliliter is equal to 0.001 L.
volume of Na2S2O3 = 1.0 mL * 0.001 L/mL = 0.001 L
moles of Na2S2O3 = 0.20 M * 0.001 L = 0.0002 moles
So, the amount of Na2S2O3 consumed for each trial is 0.0002 moles.
Now that you have the moles of Na2S2O3 consumed, you can conclude that this is also the amount of S2O3^2- consumed because there is a 1:1 molar ratio between Na2S2O3 and S2O3^2-.
Therefore, the amount of S2O3^2- consumed for each trial is 0.0002 moles.