This appears to be a limiting reactant problem.
1. Write the balanced equation
#"2NaCN" + "H"_2"SO"_4 → "Na"_2"SO"_4 + "2HCN"#
2. Calculate the moles of #"Na"_2"SO"_4# formed from the #"NaCN"#
#"Moles of Na"_2"SO"_4 = 5.93 color(red)(cancel(color(black)("g NaCN"))) × (1 color(red)(cancel(color(black)("mol NaCN"))))/(49.01 color(red)(cancel(color(black)("g NaCN")))) × (1 "mol Na"_2"SO"_4)/(2 color(red)(cancel(color(black)("mol NaCN")))) = "0.060 50 mol Na"_2"SO"_4#
3. Calculate the moles of #"Na"_2"SO"_4# formed from the #"H"_2"SO"_4#
#"Moles of Na"_2"SO"_4 = 8.45 color(red)(cancel(color(black)("g H"_2"SO"_4))) × (1 color(red)(cancel(color(black)("mol H"_2"SO"_4))))/(98.08 color(red)(cancel(color(black)("g H"_2"SO"_4)))) × (1 "mol Na"_2"SO"_4)/(1 color(red)(cancel(color(black)("mol H"_2"SO"_4)))) = "0.086 15 mol Na"_2"SO"_4#
4. Identify the limiting reactant
The limiting reactant is #"NaCN"#, because it produces the fewest moles of product.
5. Calculate the mass of #"Na"_2"SO"_4#
#"Mass of Na"_2"SO"_4 = "0.060 50" color(red)(cancel(color(black)("mol Na"_2"SO"_4))) × ("142.04 g Na"_2"SO"_4)/(1 color(red)(cancel(color(black)("mol Na"_2"SO"_4)))) = "8.59 g Na"_2"SO"_4#
