The answer is #3.0# #"g"# of #Ba(SO_4)# will form in this reaction.
Your general chemical reaction is
#Na_2SO_(4(aq)) + Ba(NO_3)_(2(aq)) -> BaSO_4(s) + 2NaNO_(3(aq))#
Notice that you've got a #1:1# mole ratio between #Ba(NO_3)_2# and #BaSO_4#; this means that for every mole of #Ba(NO_3)_2# used, 1 mole of solid will be produced.
The number of #Ba(NO_3)_2# moles can be determined using its molarity:
#C=n/V => n_(Ba(NO_3)_2) = C * V = 0.50# #"M" * 25*10^(-3)# #"L"#
#n_(Ba(NO_3)_2) = 0.013# #"moles"#
This means of course that #n_(BaSO_4) = n_(Ba(NO_3)_2) = 0.013# moles of solid will be formed. Knowing that #BaSO_4#'s molar mass is #233.3# #"g/mol"#, the mass produced will be
#m_(BaSO_4) = n_(BaSO_4) * 233.3 g/(mol) = 0.013# #"moles" * 233.3g/(mol)#
#m_(BaSO_4) = 3.0# #"g"#
The reaction's complete ionic equation is
#2Na_((aq))^(+) + SO_(4(aq))^(2-) + Ba_((aq))^(2+) + 2NO_(3(aq))^(-) -> BaSO_(4(s)) + 2Na_((aq))^(+) + 2NO_(3(aq))^(-)#
The net ionic equation is
#Ba_((aq))^(2+) + SO_(4(aq))^(2-) -> BaSO_(4(s))#
