Question 7: An interesting lab experiment is the production of insoluble Lead Io

Question

Question 7: An interesting lab experiment is the production of insoluble Lead Iodide, which is an intense yellow color (although quite highly toxic). It can be made according to the balanced reaction given below: 2 KI(aq)-Pb(NO3)2(aq) 2 K(N°3)(aq)-Pol(s) +3 points 40.52 ml of a 5.10 M solution of Potassium Iodide (KI) is mixed with 29.48 ml of a 2.40 M solution of Lead Nitrate (Pb(NO;)2). Calculate which solution will be the limiting reagent. a) b) Calculate the molarity of the resulting KNO) solution assuming the total final volume is equal o 70.00 ml.

Explanation / Answer

Ans a ) Number of moles of KI = 0.04052 x 5.10 = 0.206652 moles

Number of moles of Pb(NO3)2 = 0.02948 x 2.40 = 0.070752 moles

2 moles of KI require 1 mole of Pb(NO3)2

So 0.206652 moles would require 0.206652 / 2 = 0.103326 moles of Pb(NO3)2

So Pb(NO3)2 is the limiting reagent.

b) The number of moles of KNO3 is twice the number of moles of Pb(NO3)2

So 0.070752 x 2 = 0.141504 moles of KNO3 will be produced

Molarity = no. of moles / volume of solution in L

M = 0.141504 / 0.070

M = 2.02 M

So the molarity of KNO3 solution will be 2.02 M

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