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

668900 atoms in total

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