The mass of #"BaSO"_4"# is 32.7 g.
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)(mmmmmmmmmmmmmm) 233.39#
#color(white)(mmmmmm)"Na"_2"SO"_4 + "BaCl"_2 → "BaSO"_4 + "2NaCl"#
#"Amt/mol:"color(white)(mll)0.140color(white)(mml)0.320#
#"Divide by:"color(white)(mll)1color(white)(mmmm)1#
#"Moles rxn:"color(white)(mll)0.140color(white)(mm)0.320#
#"Moles of Na"_2"SO"_4 = 0.700 color(red)(cancel(color(black)("L Na"_2"SO"_4))) × ("0.200 mol Na"_2"SO"_4)/(1 color(red)(cancel(color(black)("L Na"_2"SO"_4)))) = "0.140 mol Na"_2"SO"_4#
#"Moles of BaCl"_2 = 0.800 color(red)(cancel(color(black)("L BaCl"_2))) × ("0.400 mol BaCl"_2)/(1 color(red)(cancel(color(black)("L BaCl"_2)))) = "0.320 mol BaCl"_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.
#"Na"_2"SO"_4# is the limiting reactant because it gives the fewest moles of reaction.
3. Calculate the mass of #"BaSO"_4"#
#"Mass of BaSO"_4 = 0.140 color(red)(cancel(color(black)("mol Na"_2"SO"_4))) × (1 color(red)(cancel(color(black)("mol BaSO"_4))))/(1 color(red)(cancel(color(black)("mol Na"_2"SO"_4)))) × ("233.39 g BaSO"_4)/(1 color(red)(cancel(color(black)("mol BaSO"_4)))) = "32.7 g BaSO"_4#
