Lead Iodide Reaction Lab
The lab today was focused on finding the ratio of reactants to products to be either 1 to 1 or 1 to 2. In our case the reactants was Lead (II) Nitrate and Potassium Iodine. These two when mixed together make Lead Iodide and Potassium Nitrate. We also had to try and find if the number of moles of Lead(II) Nitrate was the same as the final number of moles for Lead Iodine after the experiment.
Our data for the lab had pinpoint accuracy. Proved by the data table below
Trials
Volume of Pb(NO3)2
Mol Pb(NO3)2
Mol KI
Mass of Precipitate
1
1ml
0.0005
0.006
0.23g/0.0005mol
2
6ml
0.003
0.006
1.38g/0.003mol
The numbers in the data table were just simple stoichiometry. For the first trial we took the moles of Lead(II)NItrate which is 0.0005 and since the ratio from the balanced chemical equation shows that we have a 1 to 1 ratio we don’t have to put a ratio in the equation. Then we would take the molar mass of Lead Iodide. Since, that is what was mainly produced in the experiment. Then you calculate it out and repeat for the second trial except just plug in the different …show more content…
Then we heat the solution in a large beaker of water for 5 min. Then we add a drop of nitric acid, then filter and cool. Then weigh the mass, calculate the number of moles. Overall the experiment fulfilled our purposes, which was finding the ratio, which was confirmed as 1 to 1 for lead(II)Nitrate and Lead Iodine, and 1 to 2 for Potassium Iodine and Potassium Nitrate. The experiment also shows that the number of moles for Lead(II)Nitrate is exactly the same for the number of our precipitate Lead Iodide based on the data above. Though, my experiment was very accurate. There could be some errors in measurements which could cause the numbers to be off but still close. Overall the lab went very well with very few errors only thing that could be changed is cleaner
Our data for the lab had pinpoint accuracy. Proved by the data table below
Trials
Volume of Pb(NO3)2
Mol Pb(NO3)2
Mol KI
Mass of Precipitate
1
1ml
0.0005
0.006
0.23g/0.0005mol
2
6ml
0.003
0.006
1.38g/0.003mol
The numbers in the data table were just simple stoichiometry. For the first trial we took the moles of Lead(II)NItrate which is 0.0005 and since the ratio from the balanced chemical equation shows that we have a 1 to 1 ratio we don’t have to put a ratio in the equation. Then we would take the molar mass of Lead Iodide. Since, that is what was mainly produced in the experiment. Then you calculate it out and repeat for the second trial except just plug in the different …show more content…
Then we heat the solution in a large beaker of water for 5 min. Then we add a drop of nitric acid, then filter and cool. Then weigh the mass, calculate the number of moles. Overall the experiment fulfilled our purposes, which was finding the ratio, which was confirmed as 1 to 1 for lead(II)Nitrate and Lead Iodine, and 1 to 2 for Potassium Iodine and Potassium Nitrate. The experiment also shows that the number of moles for Lead(II)Nitrate is exactly the same for the number of our precipitate Lead Iodide based on the data above. Though, my experiment was very accurate. There could be some errors in measurements which could cause the numbers to be off but still close. Overall the lab went very well with very few errors only thing that could be changed is cleaner