What mass of sodium oxalate is required to ppt the Ca ion from 37.5 mL of .104M CaCl2 soln

I need to understand how to solve this problem. thanks so much

Okay, the equation of the reaction is as follows:
Na₂C₂O₄(aq)+CaCl₂(aq)
→CaC₂O₄(s)+2NaCl(aq)

Number of moles of CaCl₂
= ([CaCl₂])(volume of CaCl₂ solution)
= (0.104)(37.5x10⁻³)
= 3.90x10⁻³

According to the equation, the ratio of the number of moles of CaCl₂ required to that of Na₂C₂O₄ required is 1:1. Therefore, the number of moles of Na₂C₂O₄ required
= number of moles of CaCl₂
= 3.90x10⁻³

Relative Formula Mass(RFM) of Na₂C₂O₄
= 2(RAM of Na)+2(RAM of C)+4(RAM of O)
= 2(23.0)+2(12.0)+4(16.0)
= 134

1 mole of Na₂C₂O₄= 134g
3.90x10⁻³ moles of Na₂C₂O₄
= (134)(3.90x10⁻³)
= 0.5226g
= 0.523g correct to 3sf

Thus, 0.523g of Na₂C₂O₄ are required. I hope this helps and feel free to send me an e-mail if you have any doubts!