How do you find the limiting reagent when lead(II) nitrate solution is reacted with potassium iodide solution if 20 cm3 of each reactant is used

Question

There is 0,302 mol dm3 of lead (II) nitrate solution and 0,302 mol dm3 of potassium iodide solution

Answer 1

To find the limiting reactant, you first need the correctly balanced equation. You then need to determine which reactant will run out first (limiting) based on the stoichiometry of the balanced equation and the moles of each reactant present. In the current problem, we have..

Pb(NO₃)₂(aq) + 2KI(aq) ==> PbI₂(s) + 2KNO₃(aq) ... balanced equation

moles of Pb(NO₃)₂ present = 20 cm³ x 1L/1000 cm³ x 0.302 moles/L = 0.00604 moles Pb(NO₃)₂

moles of KI present = 20 cm³ x 1L/1000 cm³ x 0.302 moles/L = 0.00604 moles KI present

Now, note that in the balanced equation it takes TWO moles KI for each ONE mole of Pb(NO₃)₂. So, in this case, the KI is limiting and will run out first. Another way to do this, is to simply calculate the moles of product formed using each reactant. This is done as follows:

0.00604 moles Pb(NO₃)₂ x 1 mole PbI₂/1 mole Pb(NO₃)₂ = 0.00604 molers PbI₂ formed

0.00604 moles KI x 1 mole PbI₂/2 moles KI = 0.00302 moles PI₂ formed => KI is LIMITING REACTANT