Diagnostic from class
( note only 3 of the 4 select problems were worked through for your convenience)
1)
Given that 0.2 M of 0.2 L KCl reacted with 0.175 M of 0.15 L potassium chloride, find the concentration of the
resulting aqueous solution.
2KCl(aq) + Pb(NO3)2(aq)PbCl2(s) + 2KNO3(aq)
0.2 L
0.1M
0.15L
0.175 M
0.35 L
C?
NKCl= 0.1 M x 0.2L
= 0.02 mol (available)
0.02 mol KCl x 1 Pb(NO3)2 = 0.01 mol Pb(NO3)2 reacted
2 KCl
NPb(NO3)2= 0.175 M x 0.15 L
= 0. 03 mol (available)
nPb(NO3)2 (excess) since n available > n reacted
(0.02 mol remains unreacted)
KCl is the limiting reagent
0.02 mol KCl x 2 mol KNO3 = 0.02 mol KNO3 produced
2 mol KCl
[KNO3] = n/v = 0.02 mol/0.35 L= 0.057 0.06 M (1 sig digit)
2) Sodium carbonate reacts with barium nitrate. Find the minimal mass of sodium carbonate needed to precipitate all
the barium ions from solution with the reaction of 0.40 L of 0.120 M barium nitrate.
Na2CO3(aq) + Ba(NO3)2 2NaNO3(aq) + BaCO3(s)
m
0.40 L
0.12 M
nBa(NO3)2= 0.12 M x 0.4 L= 0.06 mol (available0
nNa2CO3 = 0.06 mol Ba(NO3)2 x 1 mol Na2CO3(aq)= 0.06 mol Na2CO3(aq) is needed to react with this quantity
1 mol Ba(NO3)2
mass (m) Na2CO3= 0.06 mol (105.99 g/mol)= 6.36 g 6.4 g (2 significant digits)
3) What mass of Lead (II) iodide would precipitate if 100 ml of 0.10 M Lead (II) nitrate, is added to 100 ml of 0.10 M
ammonium iodide, NH4I?
2NH4I (aq) + Pb(NO3)2(aq) PbI2(s) + 2NH4NO3
0.1L
0.10 L
0.1 M
0.1M
nNH4I = 0.1 M x 0.1 L = 0.01 mol (available) (limiting reagent)
n pb(NO3)2= 0.1 M x 0.1L = 0.01 mol (available) only 0.005 mol of which reacted (2:1 ratio, half as much reacted)
0.05 mol( excess remains unreacted)
nPbI2= 0.01 mol NH4I x 1 mol PbI2 = 0.005 mol PbI2 (produced)
2 mol NH4I
(if used mol of Pb(NO3)2 in mol ratio caclulation instead of PbI2 as above, the value would have been larger and grocely overestimated.
Mass PbI2
M= 0.005 mol x 462 g/mol PbI2= 2.3 g PbI2