Last chemistry question before final! did i do it right? How many ML of .105 M NaOH are required to

Question

neturalize exactly 14.2 mL of .041 M H3PO4

Okay I think this might be what you do but im not sure

first off I cant seem to balance this equation

H3PO4 (aq) NaOH (aq) ~~~~> H20 (l) NaPO4

do you take

14.2 mL H3PO4 x (.041 mol H3PO4/ H3PO4 liters) = .5822 mmol H3PO4

then you take

.5822=.105 M NaOH (V) = 5.54 mL NaOH

Answer 1

Moles H3PO4 = 0.0142 L x 0.041 M = 0.000582

H3PO4 + 3 NaOH >> Na3PO4 + 3 H2O

moles NaOH needed = 3 x 0.000582 =0.00175

V = 0.00175 / 0.105 =0.0166 L => 16.6 mL