How many ML of 5.6 M solution of NaCl will be necessary to make 250.0 g of AlCl3?

This reaction is not possible. The reverse reaction is favored, as sodium would rather be oxidized by aluminum, not the other way around.

And from that, I find #"1.0 L"#, or #"1000 mL"#, of #"5.6 M NaCl"(aq)# would contain the number of mols of #"NaCl"# that would be GENERATED FROM reacting #"AlCl"_3(aq)# with #"Na"(s)#.

So instead, I will solve this question:

How many mL of 5.6 M #"NaCl"# would contain the amount of mols generated by reacting 250.0 g #"AlCl"_3# with sufficient #"Na"(s)#?

#3"Na"(s) + "AlCl"_3(aq) -> 3"NaCl"(aq) + "Al"(s)#

#E_(cell)^@ = -"1.66 V" - (-"2.71 V") = "1.05 V"#
Values found here.

In this case,

#250.0 cancel("g AlCl"_3) xx "1 mol"/(133.34 cancel("g AlCl"_3)) = "1.875 mols AlCl"_3#

From the reaction coefficients, #"3 mols Na"(s)# would react with #"1 mol AlCl"_3(aq)# to form #"3 mols NaCl"(aq)#, so #"5.625 mols Na"(s)# would react with #"1.875 mols AlCl"_3# to form #"5.625 mols NaCl"(aq)#.

Therefore, we need a volume of #"5.6 M"# #"NaCl"(aq)# that contains #"5.625 mols NaCl"(aq)#. And is that not close to #"1 L"#?

#5.625 cancel("mols NaCl"(aq)) xx "1 L"/(5.6 cancel"mols") = "1.004 L"#

#~~# #color(blue)("1.0 L")#