Answer:
67.1g MgCl2
Explanation:
So first make an ICE table (Initial, Change, Equilibrium), then convert everything you have to moles. Since we know that cl2 is the limiting reagent, that means you subtract the moles Cl2 from reactant, and add it to MgCl2. Its the first substance to get completely consumed. Therefore, number of moles of Cl2 reacted = number of moles MgCl2 produced. It makes sense cause you need both of these substances together to produce MgCl2, and running out of them will stop the reaction. Lets go
20.0g x (1mol/24.305g) = 0.823mol Mg
50.0g x (1mol/70.90g) = 0.705mol Cl2
Mg + Cl2 = MgCl2
I 0.823 0.705 0
C -0.705 -0.705 +0.705
E 0.118 0 0.705
0.705mol of MgCl2 produced
0.705mol x (95.211g/mol) = 67.124g = 67.1g MgCl2 (3 s.f.)
Hope that answers your question