PAGE 307 #1-6
Pb(CH3COO)(s)  Pb2+(aq) + 2 CH3COO-(aq)
1.50 M
1.50 M
0.400 mol
0.400 mol
V = n / c = (0.400 mol Pb2+) / (1.50 mol/L Pb2+) = 0.267 L = 267 mL
6.0 g (NH4) 3PO4 x
1 mol (NH4) 3PO4 = 0.0419 mol (NH4) 3PO4
143.15 g (NH4) 3PO4
c = n / V = (0.0419 mol) / (0.300 L) = 0.14 mol/L
(NH4)3PO4(s)  3 NH4+(aq) + PO43-(aq)
0.14 M
0.42 M
0.14 M
Assume 1.0 L of each solution. They are equal volumes and this makes the math easy.
KNO3(s) 
0.12 M
0.12 mol
K +(aq)
Fe(NO3)3(s) 
0.16 M
0.16 mol
+
NO3-(aq)
0.12 M
0.12 mol
Fe3+(aq) +
3 NO3-(aq)
0.48 M
0.48 mol
n = c x V = (0.120 M)(1.0 L) = 0.12 mol
n = c x V = (0.16 M)(1.0 L) = 0.16 mol
c = n = 0.12 mol NO3- + 0.48 mol NO3- = 0. 30 mol/L
V
1.0 L +
1.0 L
Ba(NO3)2(aq) + Na2CO3(s)  BaCO3(s) + 2 NaNO3(aq)
0.120 L Ba(NO3) 2 x 0.0500 mol Ba(NO3) 2 x 1 mol Na2CO3 x
1 L Ba(NO3) 2
1 mol Ba(NO3) 2
105.99 g Na2CO3
1 mol Na2CO3(s)
= 0.64 g Na2CO3(s)
2 Al(s) + 3 CuSO4(aq)  Al2(SO4)3(aq) + 3 Cu(s)
4.88 g Cu x 1 mol Cu x
63.55 g Cu
3 mol CuSO4
3 mol Cu
x
1 L CuSO4
= 0.114 L CuSO4
0.675 mol CuSO4