Question

0.350L of 0.480M H2SO4 is mixed with 0.300L of 0.200M KOH. What concentration of sulfuric acid remains after neutralization?
H2SO4 + 2KOH = 2H2O + K2SO4

Answer 1

First work out the moles of each solution involved before the neutralization using this equation: moles = concentration x volume
H2SO4: 0.480 x 0.350 = 0.168 mol
KOH: 0.200 x 0.300 = 0.06 mol

You then deduce the moles from KOH to H2SO4 by looking at the equation. As you can see 2 moles of KOH is used to react with 1 mole of H2SO4, so we have to divide KOH's answer by 2.
(This is because it seems like KOH is the limiting reactant whereas there is an excess of H2SO4 making it the excess reactant)

0.06/2 = 0.03mol of H2SO4 reacts with 0.06mol of KOH

Now we have to work out the moles of the excess left over after the neutralization, by taking away the moles of H2SO4 at the beginning with the moles of H2SO4 that actually reacted with the KOH:

0.168 - 0.03 = 0.138 mol left over after the neutralization

Now to find the concentration of the left over H2SO4, we have to do this equation: concentration = moles/volume (we will have to add the two volumes together of each solution at the beginning for the volume)

0.138 / (0.350 + 0.300) = 0.212 M (3 sig.fig)

Therefore the concentration of the H2SO4 that remains is 0.212M
I hope this helped :)