Let's start by balancing the equation:
#AgNO_3 + H_2SO_4 -> Ag_2SO_4 + HNO_3#
#color(red)2AgNO_3 + H_2SO_4 -> Ag_2SO_4 + HNO_3#
#2AgNO_3 + H_2SO_4 -> Ag_2SO_4 + color(red)2HNO_3#
This tells us that, for every #2# moles of #AgNO_3# and #1# mole of #H_2SO_4# that reacts, #1# mole of #Ag_2SO_4# is formed.
But in the question, quantities of both #AgNO_3# and #H_2SO_4# were given. This must mean that one of them has not reacted fully and is the excess reactant.
It also means that one of them did react fully—the limiting reactant. The amount of limiting reactant is what's going to tell us how much product was formed, because it's the one that's limiting the amount of product.
To find the limiting reactant, first calculate how many moles there are for each reactant.
Moles of #AgNO_3# in #"34.7 g"# #=# #("mass of sample")/("molar mass") = 34.7/(107.87+14.01+3*16.00) = "0.204 moles"#
Moles of #H_2SO_4# in #"28.6 g"# #=# #("mass of sample")/("molar mass") = 28.6/(1.008*2+32.06+16.00*4) = "0.292 moles"#
To react with #0.204# moles of #AgNO_3#, there needs to be #0.204/2=0.102# moles of #H_2SO_4# because the mole ratio of #AgNO_3# to #H_2SO_4# is #2:1#. We have #"0.292 moles"# of #H_2SO_4#, therefore #H_2SO_4# cannot be the limiting reactant.
To react with #0.292# moles of #H_2SO_4#, there needs to be #0.292*2=0.584# moles of #AgNO_3# because the mole ratio of #AgNO_3# to #H_2SO_4# is #2:1#. We have #0.204# moles of #AgNO_3#, therefore #AgNO_3# is the limiting reactant.
#2AgNO_3 + H_2SO_4 -> Ag_2SO_4 + 2HNO_3#
Every #2# moles of #AgNO_3# corresponds to #1# mole of #Ag_2SO_4#.
We have #0.204# moles of #AgNO_3#, therefore we also have #0.204/2=0.102# moles of #Ag_2SO_4#.
Finally, to find the mass of #Ag_2SO_4#, we multiply the number of moles by the mass per mole: #0.102 * "molar mass" = 0.102 * (2*107.87 + 32.06 + 4*16.00) = "31.8 g"#.
