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"NaHCO3+HC2H3O2→NaC2H3O2+CO2+H2O
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).
