I need to find the % yield for the making of myristic acid from a reaction of from trimyristin with other chemicals. 2mL NaOH, 2mL C2H5OH, 10mL HCl,...

I need to find the % yield for the making of myristic acid from a reaction of from trimyristin with other chemicals.

2mL NaOH, 2mL C2H5OH, 10mL HCl, 0.2g trimyristin are the reactants

0.04g found of myristic acid

My first problem is finding the moles of myristic acid. In the balanced equation everything is 1mol but NaOH and the product myristic acid which have 3 moles.These are the claculation of the others. The molar mass for myristic acid is 228.37g/mol

NaOH 2mL * 2.13g/mL* 3mol/(39.99g/mol) = 0.3196mol

C2H5OH 2mL* 0.789g/mL* 1mol/(46.07g/mol) = 0.03425mol

HCl 10mL* 1.048g/mL* 1/(36.46g/mol) = 0.2874mol

Myristic acid .04g * 3mol/((3*228.37g/mol)) = 1.751*10-4

trimyristin 0.2g * 1mol/(723.16g/mol) = 2.766*10-4

I know the next steps are then figure out the limiting reagent,determine the theoretical number of moles of product, determine the actual number of moles of product, then the % yield equation can be used.

Please help me out with the steps