In a reaction involving sodium metal and chlorine gas, the amount of sodium metal required to react with 0.950 grams of chlorine gas would be 0.62 grams
<h3>Stoichiometric calculation</h3>
From the balanced equation of the reaction:
2Na (s) + Cl2 (g) ---------> 2NaCl (s)
The mole ratio of Na to Cl2 is 2:1.
Mole of 0.950 grams of chlorine gas = 0.950/70.906
= 0.0134 moles
Equivalent mole of Na = 0.0134 x 2 = 0.0268 moles
Mass of 0.0268 moles Na = 0.0268 x 22.99
= 0.62 grams
More on stoichiometric calculations can be found here: brainly.com/question/8062886
