Question

I'm having a hard time doing the prelab work for my Mass-Mass Stoichiometry Lab.
The balance equation is CaCl2 (aq) + Na2CO3 (q) -> CaCO3(s) + 2NaCl(aq)
Pre-Lab work:
1b) Calculate the mass of CaCl2 that will be needed to react with the 4.00 grams of Na2CO3•H2O. (Note: The •H2O is included in calculating molecular mass of Na2CO3.H2O to get the mass of CaCl2 needed) Answer:3.58 g CaCl
***Include your computations of the amount of water that should be used to dissolve the reactants.* (This is where I need help.)

Answer 1

This is the lab attachment. For the prelab, I just had to do 1a and 1b, but it also asked for the minimum amount of water needed, and that's where I'm struggling.
CaCl2=3.58 g
NaCO3=4g (provided)
CaCO3=3.23 g
2NaCl=3.77g

Answer 2

\[1 mol Na2CO3•H2O \times )1 mol \div 124.01 g) \times (1 mol H2O \div 1 mol) \times (18.02g H2O \div 1 mol H2O) =0.58 g H2O\]

So I think my mass of water would need to be 0.58 g or when converted to mL, 0.74 mL of H2O. I'm still not sure about this answer, but it's what I've been able to figure out.