So we are told that sodium and diatomic chlorine react to form sodium chloride NaCl. So let's create and balance the equation:
→

So now that we have the molar ratio of 2:1:2, we can use this ratio to help solve the rest.
We are given 117 grams of NaCl, so we need to find out how many moles this is. We add the molar weight of sodium (22.99g/mol) and chlorine (35.45g/mol) to find the molar weight of the compound:

So then we solve for how many moles 117g is:

So since we know that 2 mol of NaCl was produced in the reaction, we know that this follows the ratio of the balanced equation. Therefore, only 1 mole of
was used in the reaction.
So we find the mass the same way - but there are two atoms of chlorine. So we are going to take the atomic weight of chlorine (35.45g) from the periodic table and multiply by 2:

Therefore, 70.9g of chloride was needed for the reaction.