To solve this question, we need to find the limiting reactant first. We can do this by calculating the number of moles of each reactant:
moles of CuCl2 = 342 g / 134.45 g/mol = 2.547 moles
moles of NaNO3 = 217 g / 84.99 g/mol = 2.553 moles
Since NaNO3 has more moles, it is in excess and CuCl2 is the limiting reactant.
We can use the stoichiometry of the balanced chemical equation to find the number of moles of NaCl produced:
CuCl2 + 2NaNO3 → 2NaCl + Cu(NO3)2
From the equation, we can see that 1 mole of CuCl2 reacts with 2 moles of NaCl. Therefore, 2.547 moles of CuCl2 will produce:
2.547 moles CuCl2 x (2 moles NaCl / 1 mole CuCl2) = 5.094 moles NaCl
Therefore, 5.094 moles of NaCl will be produced.