This is really a two-part question:
- How much of the #"H"_2"SO"_4# is neutralized?
- What is the concentration of the remaining #"H"_2"SO"_4#?
1. How much of the #"H"_2"SO"_4# is neutralized?
This is a volume-moles stoichiometry problem.
The steps are:
a. Write the balanced equation.
b. Use the molarity of #"KOH"# to convert volume of #"KOH"# to moles of #"KOH"#.
c. Use the molar ratio from the balanced equation to convert moles of #"KOH"# to moles of #"H"_2"SO"_4# that reacted.
d. Use the molarity to convert the initial volume of #"H"_2"SO"_4# to moles of #"H"_2"SO"_4#.
e. Calculate the moles of excess #"H"_2"SO"_4#.
a. Write the balanced equation:
#"2KOH" + "H"_2"SO"_4 → "K"_2"SO"_4 + "2H"_2"O"#
b. Calculate the moles of #"KOH"#.
#"Moles of KOH" = 0.800 color(red)(cancel(color(black)("L KOH"))) × "0.280 mol KOH"/(1 color(red)(cancel(color(black)("L KOH")))) = "0.224 mol KOH"#
c. Calculate the moles of #"H"_2"SO"_4# that reacted
The molar ratio of #"H"_2"SO"_4:"KOH"# is #"1 mol H"_2"SO"_4:"2 mol KOH"#.
∴ #"Moles of H"_2"SO"_4 = 0.224 color(red)(cancel(color(black)("mol KOH"))) × (1 "mol H"_2"SO"_4)/(2 color(red)(cancel(color(black)("mol KOH")))) = "0.112 mol H"_2"SO"_4#
2. What is the concentration of the remaining #"H"_2"SO"_4#?
a. Calculate the initial moles of #"H"_2"SO"_4#.
#"Moles of H"_2"SO"_4 = 0.850 color(red)(cancel(color(black)("L H"_2"SO"_4))) × ("0.490 mol H"_2"SO"_4)/(1 color(red)(cancel(color(black)("L H"_2"SO"_4)))) = "0.4165 mol"#
e. Calculate the moles of #"H"_2"SO"_4# remaining
#"Moles remaining" = "initial moles – moles reacted" = "0.4165 mol – 0.112 mol" = "0.3045 mol"#
e. Calculate the molarity of the excess #"H"_2"SO"_4#
#"Molarity" = "moles"/"litres"#
#"Litres" = "Volume of KOH + Volume of H"_2"SO"_4 = "0.800 L + 0.850 L" = "1.650 L"#
∴ #"Molarity" = "0.3045 mol"/"1.650 L" = "0.185 mol/L"#
