IB Chemistry Paper 2 HL 2024 - (Mock Exam /Prediction Questions)

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I want to revise the answer of number 1 b(ii):
The equation should be:
3 Ca(ClO3)2 + 2 Na3PO4 produces Ca3(PO4)3 + 3 NaCl
since mol of Ca(ClO3)2 is 0.01713 in the Q1a (this is for 3 Ca(ClO3)2), so for 1 Ca(ClO3)2 is 1/3 x 0.01713 = 0.00571.
mol of Na3PO4 is C x V (from Qb(i)) = 0.075 x 0.2= 0.15 (this for 2 Na3PO4), thus for 1 Na3PO4, 1/2 x 0.15 = 0.0075
Thus n Na3PO4 is bigger than n Ca(ClO3)2, thus Ca(ClO3)2 is the limiting reactant.

[RinkiKumari-edbn]

4:11 how did you calculate the mole of Na3PO4 here as when we divide the moles of CaClo3 with 3 we get 0.00571, isnt that correct??? also how did you get here the value 0.0075?? im so confused