Question in the description, can you please explain how you got the answer, thank you

Question

What is the greatest amount of copper IV nitrate in moles that can be made with 24.3 mol copper II chloride and 20.9 mol sodium nitrate? Which reactant is limiting? How many moles of the excess reactant will be left after the reaction?

CuCl₂  + 2 NaNO₃  Cu(NO₃)₂ + 2 NaCl

Answer 1

24.3 mol CuCl2 : 20.9 mol NaNO3

You see that for each mol of CuCl2 you need 2 mol of NaNO3

Therefore 24.3 mol CuCl2 will react 1:2 with 48.6 mol

OR you could have started the other direction

For every 2 mol NaNO3, you will react 1 mol CuCl2

So 20.9 mol NaNO3 will react (theor.100%) with 10.45 mol CuCl2 (ignoring s.f.)

Hence we see there is excess CuCl2 (NaNO3 limiting reagant)

The NaNO3 if reacted completely would consume ~10.5 mol CuCl2,

This leaves 24.3 mol CuCl2 - 10.5 mol CuCl2, or 13.8 moles if my quick mental calc is correct.

K? K.