2Na + Cl2 → 2NaCI How many grams of NaCl will be produced from 33.0 g of Na and 34.0 g of Cl.?

56.0 grams NaCl

Explanation:

First, determine the limiting reactant by converting either one of the reactants' available quantities into the other reactant using the balanced equation. I picked 33.0 g Na:

33.0 g Na x 1 mol Na/22.99 g Na = 1.435 mol Na

1.435 mol Na x 1 mol Cl2/2 mol Na = 0.718 mol Cl2

0.718 mol Cl2 x 70.906 g Cl2/1 mol Cl2 = 50.89 g Cl2

As we can see, we only have 34.0 g of Cl2 to react with, though 33.0 g of Na can react with 50.89 g of Cl2. This means that Cl2 is our limiting reactant.

Now we can work off of the limiting reactant and stoichiometrically determine the grams of NaCl produced:

34.0 g Cl2 x 1 mol Cl2/70.906 g Cl2 = 0.4795 mol Cl2

0.4795 mol Cl2 x 2 mol NaCl/1 mol Cl2 = 0.9590 mol NaCl

0.9590 mol NaCl x 58.44 g NaCl/1 mol NaCl = 56.04 g NaCl

Round 56.04 g NaCl to three significant figures -> 56.0 g NaCl

it's c. elements are placed in rows based on how may valencies electrons have.