Answer
a)$BaCl_2(aq)+2\ AgNO_3(aq)\rightarrow 2\ AgCl(s)+ Ba(NO_3)_2(aq)$
b)
$0.254\ g\ AgNO3$
$0.215\ g\ AgCl$
Work Step by Step
a)$BaCl_2(aq)+2\ AgNO_3(aq)\rightarrow 2\ AgCl(s)+ Ba(NO_3)_2(aq)$
b) Number of moles of barium chloride:
$0.156\ g\div 208.233\ g/mol=0.000749\ mol$
From stoichiometry:
$0.000749\ mol\times2/1\times 169.873\ g/mol\ AgNO3=0.254\ g\ AgNO3$
$0.000749\ mol\times2/1\times 143.321\ g/mol\ AgNO3=0.215\ g\ AgCl$