Two days ago, we mentioned that Mole Calculations is the number 1 killer topic in O Level Chemistry, and many students are not doing well for it. In fact, a large number of students skip the calculation-based questions when they see it on examination, fearing that they spend alot of time on that question, and yet could not answer it.
Over the years, this topic is often voted the “Hot Favorite” on students’ challenging topics. Some have
- problems understanding what the question is asking for,
- do not know where to start,
- totally blank when they read the question,
- too confused with the formulaes to use…WHY?
Let’s check out a typical exam question on Mole Calculations:
Question:
The following equation represents the precipitation reaction between lead (II) nitrate solution and potassium iodide solution.
aPb(NO3)2(aq) + bKI(aq) –> cPbI2(s) + dKNO3(aq)
In an experiment, 20.0g of lead(II) nitrate and 20.0g of potassium iodide were dissolved separately to form 250cm3 of lead (II) nitrate solution and 250cm3 of potassium iodide solution.
(a) Determine a, b, c and d. (2 marks)
(b) Using suitable calculation, identify the limiting reactant. (2 marks)
(c) What is the maximum mass of lead (II) iodide (PbI2) that can be obtained from the experiment? (2 marks)
(d) What is the concentration of potassium nitrate in the reaction mixture when the reaction is completed? Give your answers in mol/dm3. (2 marks)
(e) Write an Ionic Equation for the above reaction and identify the spectator ion(s). (2 marks)
PS: Do let me know your suggested answers in the “Comments Section” below.
Answer:
a) a = 1 , b = 2, c = 1 and d = 2 b) KI is the Limiting Reactant.
c) Maximum mass of PbI2 that can be obtained = 27.8g (to 3 sf)
d) Concentration of KNO3 = 0.241 mol/dm3
e) Ionic Equation: Pb2+(aq) + 2I-(aq) ? PbI2(s)
Spectator Ions: NO3-(aq) and K+(aq)
PPS: This is only one type of mole calculation questions. There are many other types of questions that combine several topics together.
can u pls expalin why KI is the limiting factor???
For question part d, just calculate the number of moles of KNO3 using mole ratio of the equation and divide by the the total volume of aqueous solution which is 500cm3.
remember to convert it to dm3.
No. of Moles of Pb(NO3)2 = 20 / (207 + 2x(14+16×3)) = 0.0604229
No. of Moles of KI = 20 / (39 +127) = 0.1204819
Based on the equation, by right, 0.0604559 mol of Pb(NO3)2 require 0.1208458 mol of KI. Since, no of moles of KI is 0.1204819 only, KI is the limiting reactant.
why isnt the limiting reagent is Pb(NO3)? please reply asapppppp
wow some good questions, could u please send me the working out for question (d) What is the concentration of potassium nitrate in the reaction mixture when the reaction is completed? Give your answers in mol/dm3. (2 marks)! because i dont know where i am going wrong! thank you
x
Please explain why the limiting reactant is KI and not Pb(NO3)2. Thanks