a,250ml=0.25l
$C_{M}$= $\frac{n}{V}$
hay 2= $\frac{n}{0.25}$
=>nBa(OH)2=2*0.25=0.5(mol)
=>mBa(OH)2=0.5*(137+16*2+2)=85.5(g)
b,80ml=0.08l
$C_{M}$= $\frac{n}{V}$
hay 0.15=$\frac{n}{0.08}$
=>nFeCl3=0.15*0.08=0.012(mol)
=>mFeCl3=0.012*(56+35.5*3)=1.95(g)