Asked by mel
                I balanced 
Li(s) + Cl2(g)= LiCl(s) to
2Li(s) + Cl2(g)= 2LiCl(s)
Now the question is, if you have 39.8g Li, how many moles of LiCl would you have?
I am stuck on this part. Any help would be appreciated.
            
        Li(s) + Cl2(g)= LiCl(s) to
2Li(s) + Cl2(g)= 2LiCl(s)
Now the question is, if you have 39.8g Li, how many moles of LiCl would you have?
I am stuck on this part. Any help would be appreciated.
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