When 2.3 g of na (s) reacts with excess cl2, how many grams of nacl will be produced?

The reaction between Na and Cl2 is as follows;

2Na + Cl2 - > 2NaCl

Stoichiometry of Na to NaCl is 2:2

Since Cl2 is provided in excess, Na is the limiting reactant. Therefore amount of product formed depends on amount of Na present.

Mass of Na reacted = 2.3 g

Number of Na moles reacted = 2.3 g/23 g / mol = 0.1 mol

Therefore number of NaCl moles produced = 0.1 mol

The mass of NaCl produced = 0.1 mol x 58.5 g/mol = 5.85 g