Calculate how many grams of the first reactant are necessary to completely react with 55.8g of the second reactant.

Question

Na2CO3 + CuCl2 → CuCO3 + 2 NaCl

Answer 1

It's just another stoichiometry problem.
mols CuCl2 = grams/molar mass = approx 0.4 but you need to be more accurate than that.
Then 0.4 mol CuCl2 x (1 mol Na2CO3/1 mol CuCl2) = 0.4 x 1/1 = 0.4 mol Na2CO3. To convert that to grams you have
g = mols Na2CO3 x molar mass Na3CO3.

Answer 2

55.8gCuCl2 x (1molCuCl2/134.45gCuCl2) x (1molNa2CO3/1molCuCl2) x (105.99gNaCO3/molNa2CO3) =44.0gNa2CO3
I Think

Answer 3

Yes, I get 43.988 g which rounds to 44.0 to 3 s.f.

Answer 4

13.8g