Lead nitrate (Pb(NO3)2) and potassium iodide (KI) react in aqueous solution to form a yellow precipi — Physical Chemistry — Kinetics Chemistry Question
Lead iodide
Lead nitrate (Pb(NO3)2) and potassium iodide (KI) react in aqueous solution to form a yellow precipitate of lead iodide (PbI2). In one series of experiments the masses of the two reactants were varied, but the total mass of the two was held constant at 5.000 g. The lead iodide formed was filtered from solution, washed and dried. Data for a series of reactions are given below, together with a blank graph.
Experiment | Mass of lead nitrate (g) | Mass of lead iodide (g)
1 | 0.500 | 0.692
2 | 1.000 | 1.388
3 | 1.500 | 2.093
4 | 3.000 | 2.778
5 | 4.000 | 1.391
[VISUAL]
Complete the graph; that is, plot the data and draw the approximate curve(s) connecting the data points. Determine graphically what the maximum mass of precipitate is that can be obtained?
Model Answer
The graph obtained is one of two straight lines, meeting at a peak of about 2.50 g of Pb(NO3)2.
Data according to the reaction:
KI(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + PbI2(s)
Mass of Pb(NO3)2 (g) | Mass of PbI2 (g)
0.500 | 0.696
1.000 | 1.392
1.500 | 2.088
4.000 (1.000 g KI) | 1.389
3.000 (2.000 g KI) | 2.778
[VISUAL]
Write the balanced equation for the reaction and use it to calculate the maximum mass of PbI2 and the corresponding amount of Pb(NO3)2.
Model Answer
The total quantity of reactant is limited to 5.000 g. If either reactant is in excess, the amount in excess will be “wasted”, because it cannot be used to form product. Thus, we obtain the maximum amount of product when neither reactant is in excess; there is a stoichiometric amount of each.
The balanced chemical equation for this reaction:
2 KI(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + PbI2(s)
shows that stoichiometric quantities are two moles of KI (M = 166.00 g mol-1) for each mole of Pb(NO3)2 (M = 331.21 g mol-1). If we have 5.000 g total, we can let the mass of KI equal to x g, so that the mass of Pb(NO3)2 = (5.000 – x) g. Then we have:
n(KI) = x / 166.00 mol
n(Pb(NO3)2) = (5.000 - x) / 331.21 mol
At the point of stoichiometric balance, n(KI) = 2 n(Pb(NO3)2):
x / 166.00 = 2 * (5.000 - x) / 331.21
x = 2.503 g
n(Pb(NO3)2) = (5.000 - 2.503) / 331.21 = 0.0075 mol = n(PbI2)
Maximum mass PbI2 = n(PbI2) * M(PbI2) = 0.0075 mol * 461.0 g mol-1 = 3.476 g