a)

Ba(OH)_2+K_2SO_4->BaSO_4+2KOH

500ml=0,5l;250ml=0,25l

n_{Ba(OH)_2}=0,5.0,5=0,25(mol)

n_{K_2SO_4}=0,25.0,5=0,125(mol)

b)

Do 0,25>0,125->Ba(OH)_2 dư.

Theo PT: n_{KOH}=2n_{K_2SO_4}=0,25(mol)

->m_{KOH}=0,25.56=14(g)

Theo PT: n_{Ba(OH)_2\ pu}=n_{K_2SO_4}=0,125(mol)

->n_{Ba(OH)_2\ du}=0,25-0,125=0,125(mol)

->m_{Ba(OH)_2\ du}=0,125.171=21,375(g)

c)

V_{dd\ spu}=0,5+0,25=0,75(l)

->C_{M\ KOH}={0,25}/{0,75}=1/3M

        C_{M\ Ba(OH)_2\ du}={0,125}/{0,75}=1/6M