Answer:
moles SO42- = 0.0500 L x 0.20 M=0.010
moles Ba2+ = 0.0500 L x 0.10 M = 0.0050
Ba2+ + SO42- = BaSO4 (s)
moles SO42- in excess = 0.010 - 0.0050=0.0050
total volume = 100 mL = 0.100 L
[SO42-]= 0.0050/0.100= 0.050 M
[Na2SO4] = 2 /2 = 1 M
moles Na2SO4 = 2 M x 0.500 L = 0.500
mass Na2SO4 = 0.500 x 142 g/mol=71.0 g
moles MgBr2 = 46 /184 =0.25
moles Br- = 0.25 x 2 = 0.50
[Br-]= 0.50 / 0.50 L = 1 M
