xLi + Cl2 ---> xLiCl
You have 2 Cl, so at the end you need the same 2. If the resulting component is LiCl, you might have 2LiCl, so you spend the 2 initial Cl
Because of this, the result have 2 Li, so at the begining you have to have 2 Li.
D. 2
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