how do you find the limiting reagent of .35 grams of potassium iodide and .45 grams of lead (II) nitrate

Question

i have the correct balanced equation and the molarity of potassium iodide. i also have the mass of the precipitate. 

Answer 1

To answer this question, we'll need the balanced equation:

 

2KI + Pb(NO₃)₂ --> 2KNO₃ + PbI₂

 

First, we determine the moles of each reactant present, by dividing the mass given by the molar mass for each compound:

 

0.35 g KI x (1moles/166.0g) = 0.0021 moles KI

 

0.45 g Pb(NO₃)₂ x (1 mole/331.2 g) = 0.0014 moles Pb(NO₃)₂

 

From there, we determine how many moles of the other reactant would be required to completely react with the number of moles of either given reactant by multiplying by the mole ratio from the balanced chemical equation:

 

 

0.0014 moles Pb(NO3)2 x (2 moles KI/1 mole Pb(NO3)2) = 0.0028 moles KI

 

This tells us we would need 0.0028 moles of Ki to completely react with the 0.0014 moles of Pb(NO3)2 that we have. By we only have 0.0021 moles of KI, which means we will run out before all of the Pb(NO3)2 reacts, so KI is limiting. 

 

We could have also started with moles present of KI:

 

0.0021 moles KI x (1 mole Pb(NO3)2/2 moles KI) = 0.00105 moles of Pb(NO3)2

 

This tells us that we would need 0.00105 moles of lead (II) nitrate to completely react with the 0.0021 moles of KI that we have. But we have 0.0014 moles of Pb(NO3)2, which means we have excess and will not run out. (So again, the other reactant KI is limiting).