This is a limiting reactant problem.
We know that we will need a balanced equation with masses, molar masses, and moles of the compounds involved.
#color(blue)(bb"A."color(white)(l) "Identify the limiting reactant")#
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) 98.08 color(white)(mmll)40.00color(white)(mmll)142.04#
#color(white)(mmmmmmm) "H"_2"SO"_4 + "2NaOH" → "Na"_2"SO"_4 + "2H"_2"O"#
#"Mass/g:"color(white)(mmmll)9.65color(white)(mmm)6.10#
#"Amt/mol:"color(white)(ml)"0.098 38"color(white)(mll)0.1525#
#"Divide by:"color(white)(mmm)1color(white)(mmmmll)2#
#"Moles rxn:"color(white)(m)"0.098 38"color(white)(mll)"0.076 25"#
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.
#"NaOH"# is the limiting reactant because it gives the fewest moles of reaction.
#color(blue)(bb"B."color(white)(l) "Calculate the theoretical yield of Na"_2"SO"_4)#
2. Calculate the theoretical moles of #"Na"_2"SO"_4#
#"Theoretical yield" = 0.1525 color(red)(cancel(color(black)("mol NaOH"))) × ("1 mol Na"_2"SO"_4)/(2 color(red)(cancel(color(black)("mol NaOH")))) = "0.076 25 mol Na"_2"SO"_4#
3. Calculate the theoretical yield of #"Na"_2"SO"_4#
#"Theoretical yield" = "0.076 25" 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)))) = "10.83 g Na"_2"SO"_4#
#color(blue)(bb"C."color(white)(l) "Calculate the mass of excess reagent")#
4. Calculate the moles of #"H"_2"SO"_4# reacted.
#"Moles of H"_2"SO"_4 = 0.1525 color(red)(cancel(color(black)("mol NaOH"))) × ("1 mol H"_2"SO"_4)/(2 color(red)(cancel(color(black)("mol NaOH")))) = "0.076 25 mol H"_2"SO"_4#
5. Calculate the mass of #"H"_2"SO"_4# reacted.
#"Mass of H"_2"SO"_4 = "0.076 25" color(red)(cancel(color(black)("mol H"_2"SO"_4))) × ("98.08 g H"_2"SO"_4)/(1 color(red)(cancel(color(black)("mol H"_2"SO"_4)))) = "7.479 g H"_2"SO"_4#
6. Calculate the mass of #"H"_2"SO"_4# remaining.
#"Mass of H"_2"SO"_4 = "9.65 g - 7.479 g" = "2.15 g H"_2"SO"_4#
