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
Stoichiometric calculation
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
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