We are given the amounts of three reactants, so this is a **limiting reactant** problem.

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.

**1. Gather all the information** in one place with molar masses above the formulas and everything else below the formulas.

#M_r:color(white)(mmmmm) 78.04color(white)(mm)105.99color(white)(mm)64.06color(white)(mmll)158.11#

#color(white)(mmmmmmm)"2Na"_2"S" + "Na"_2"CO"_3 + "4SO"_2 → "3Na"_2"S"_2"O"_3 + "CO"_2#

#"Mass/g":color(white)(mmml)200color(white)(mmmll)200color(white)(mmll)200#

#"Amt/mol:"color(white)(mml)2.563color(white)(mmll)1.887color(white)(mm)3.122#

#"Divide by:"color(white)(mmm)2color(white)(mmmmm)1color(white)(mmmm)4#

#"Moles rxn:"color(white)(mm)1.281color(white)(mmll)1.887color(white)(mll)0.7805#

#"Moles of Na"_2"S" = 200 color(red)(cancel(color(black)("g Na"_2"S"))) × ("1 mol Na"_2"S")/(78.04 color(red)(cancel(color(black)("g Na"_2"S")))) = "2.563 mol Na"_2"S"#

#"Moles of Na"_2"CO"_3 =200 color(red)(cancel(color(black)("g Na"_2"CO"_3))) × ("1 mol Na"_2"CO"_3)/(105.99 color(red)(cancel(color(black)("g Na"_2"CO"_3)))) = "1.887 mol Na"_2"CO"_3#

#"Moles of SO"_2 = 200 color(red)(cancel(color(black)("g SO"_2))) × ("1 mol SO"_2)/(64.06 color(red)(cancel(color(black)("g SO"_2)))) = "0.7805 mol SO"_2#

**2. Identify the limiting reactant**

An easy way to identify the limiting reactant is to calculate the "moles of reaction" each will give:

You divide the moles of each reactant by its corresponding coefficient in the balanced equation.

I did that for you in the table above.

#"SO"_2# is the limiting reactant because it gives the fewest moles of reaction.

**3. Calculate the theoretical moles of #"Na"_2"S"_2"O"_3#**

#"Theoretical moles" = 3.122 color(red)(cancel(color(black)("mol SO"_2))) × ("3 mol Na"_2"S"_2"O"_3)/(4 color(red)(cancel(color(black)("mol SO"_2)))) = "2.342 mol Na"_2"S"_2"O"_3"#

**4. Calculate the theoretical yield of #"Na"_2"S"_2"O"_3#**

#"Theoretical yield" = 2.342 color(red)(cancel(color(black)("mol Na"_2"S"_2"O"_3))) × ("158.11 g Na"_2"S"_2"O"_3)/(1 color(red)(cancel(color(black)("mol Na"_2"S"_2"O"_3)))) = "370 g Na"_2"S"_2"O"_3#