Standardization and Titration?

Question

Here's the info I have:
.4063 g KHP
Titrated 17.93 mL of NaOH into the KHP solution
Mole ratio KHP:NaOH, 1:1

10 mL H2SO4
Titrated 26.39 mL NaOH into the acid
Mole ratio NaOH:H2SO4 is 2:1

Please help me!

Part 1 - Standardization
1. Calculate the moles of KHP from mass weighed out.
2. Calculate the moles of NaOH that would react with the KHP.
3. Calculate the concentration of NaOH using titration volume.

Part 2 - Titration of Acid
1. Calculate the moles of NaOH delivered from buret
2. Calculate the moles of acid that would react with the NaOH
3. Calculate the concentration of acid using sample volume.

Answer 1

0.4063 g / 204.22 g/mole = 0.001990 moles KHP
moles of NaOH = 0.001990 (1.990 millimoles)
mL NaOH x Molarity NaOH = 1.990 millimoles NaOH
17.93 x M = 1.990
Molarity of NaOH = 0.1110

26.39 mL NaOH x 0.1110 M = 2.928 millimoles NaOH
2.928 mmoles of NaOH = 1.464 mmoles of H2SO4
(2 moles NaOH to 1 mole of H2SO4)
1.464x10^-3 moles x 98 g/mole = 0.1435 grams H2SO4
1.43% (w/v) H2SO4

Answer 2

Molecular weight KHP = 204.22

Let NaOH solution be called SHS

0.4063gKHP/17.93mLSHS x 1molKHP/204.2gKHP x 1molNaOH/1molKHP x 1000mLSHS/1LSHS = 0.1110 mole NaOH per 1L solution

0.4063gKHP x 1molKHP/204.2gKHP x 1molNaOH/1molKHP = 0.001990 mole NaOH = 0.001990 mole KHP

Maybe you can work out the second half.