If 2.0 g of copper(II) chloride react with excess sodium nitrate?

This is not a double replacement reaction . No NaCl is produced. All that you have on mixing these two solutions is a solution containing a mixture of Cu2+ , Cl- , Na+ and NO3- ions.
Equation:
CuCl2(aq) + NaNO3(aq) → No reaction.

No reaction occurs between copper(II) chloride and sodium nitration.

If there were reaction, the equation should be CuCl₂ + 2NaNO₃ → Cu(NO₃)₂ + 2NaCl
Mole ratio CuCl₂ : NaCl = 1 : 2

Molar mass of CuCl₂ = (63.5 + 35.5×2) g/mol = 134.5 g/mol
No. of moles of CuCl₂ reacted = (2.0 g) / (134.5 g/mol) = 0.0149 mol
No. of moles of NaCl formed = (0.0149 mol) × 2 = 0.0298 mol

Molar mass of NaCl = (23.0 + 35.5) g/mol = 58.5 g/mol
Mass of NaCl formed = (0.0298 mol) × (58.5 g/mol) = 1.7 g (to 2 sig. fig.)

copper(II) chloride and sodium nitrate .... what the heck.....

Trevor is dead on. There is NO reaction. All of the answers with masses of products are nonsense. When the two solutions combined you get the same four ions you started with, and no precipitates. There is nothing to compute the mass of.

CuCl2 + 2NaNO3 = 2NaCl + Cu(NO3)2
The molar ratios are 1:2::2:1
So 1 mole of CuCl2 produces 2 moles NaCl
Hence moles CuCl2 = 2.0/(63.5 + (2 x 35.5)) = 2 / 134.5 = 0.01487 (Equivalent to '1')
moles NaCl = 0.01487 x 2 = 0.02974 (Equivalent to '2').
mass(NaCl) = 0.02974 = mass(g) / (23 + 35.5) = >
mass(NaCl) = 0.02974 x 58.8 = 1.7398 g

I'll write the balanced equation afterward, but it is definitely NOT needed, to answer the question:

(2.0 g)(58.44 g/mol)/(134.45 g/mol) = 0.87 grams.
As ALL the chlorine will be used in the reaction, one need account only for the loss of mass that ensues from replacing copper with sodium.

Equation:
CuCl2 + 2NaNO3 --> 2NaCl + Cu(NO3)2