Question

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.

Answer 1

Divide 39.8 g by the atomic weight of Li, 6.94 g/mole. That gives you 5.735 moles of lithium reactant.

You form the same number of moles of LiCl product.

Answer 2

Thank you!