The equation for the reaction is
#"NaHCO"_3 + "HC"_2"H"_3"O"_2 → "NaC"_2"H"_3"O"_2 + "CO"_2 + "H"_2"O"#
Step 1. Calculate the moles of sodium acetate
#"Moles of NaC"_2"H"_3"O"_2 = 2.5 color(red)(cancel(color(black)("g NaC"_2"H"_3"O"_2))) × ("1 mol NaC"_2"H"_3"O"_2)/(82.03 color(red)(cancel(color(black)("g NaC"_2"H"_3"O"_2)))) = "0.0305 mol NaC"_2"H"_3"O"_2#
Step 2. Calculate the mass of baking soda required
#"Moles of NaHCO"_3 = 0.0305 color(red)(cancel(color(black)("mol NaC"_2"H"_3"O"_2))) × ("1 mol NaHCO"_3)/(1 color(red)(cancel(color(black)("mol NaC"_2"H"_3"O"_2)))) = "0.0305 mol NaHCO"_3#
#"Mass of NaHCO"_3 = 0.0305 color(red)(cancel(color(black)("mol NaHCO"_3))) × ("84.01 g NaHCO"_3)/(1 color(red)(cancel(color(black)("mol NaHCO"_3)))) = "2.6 g NaHCO"_3 #
Step 3. Calculate the volume of acetic acid
#"Moles of HC"_2"H"_3"O"_2 = 0.0305 color(red)(cancel(color(black)("mol NaC"_2"H"_3"O"_2))) × ("1 mol HC"_2"H"_3"O"_2)/(1 color(red)(cancel(color(black)("mol NaC"_2"H"_3"O"_2)))) = "0.0305 mol HC"_2"H"_3"O"_2#
# "Volume of HC"_2"H"_3"O"_2 = 0.0305 color(red)(cancel(color(black)("mol HC"_2"H"_3"O"_2))) × ("1 L HC"_2"H"_3"O"_2)/(0.87 color(red)(cancel(color(black)("mol HC"_2"H"_3"O"_2)))) = "0.035 L HC"_2"H"_3"O"_2 = "35 mL HC"_2"H"_3"O"_2#
You need either #"2.6 g of NaHCO"_3# or #"35 mL of HC"_2"H"_3"O"_2#, whichever is less (the limiting reactant).
