If the chemist has 35.0 g of Na, then 26.94 grams of chlorine is required to react completely with the sodium.
Number of moles of any substance can be calculated as [tex]${\rm{n = }}\frac{{\rm{W}}}{{\rm{M}}}$[/tex].
Where, W = given mass
M = molar mass
Balanced equation for the formation of NaCl can be written as:
[tex]${\rm{2N}}{{\rm{a}}_{{\rm{(s)}}}}{\rm{ + C}}{{\rm{l}}_{\rm{2}}}_{{\rm{(g)}}} \to {\rm{2NaC}}{{\rm{l}}_{{\rm{(s)}}}}$[/tex]
Here the ration of Na and Cl is 2:1.
No. of moles of Na = 35/23 = 1.52 mole
If 2 moles of Na is consumed by 1 mole of Cl, then 1 mole of Na is consumed by half mole of Cl.
Mole of Cl required to consume 1 mole of Na = 1.52/2 = 0.760 mole.
Now used mass of Cl can be find out by using the formula of moles as follow:
W = 35.45 × 0.760 = 26.94 grams
Hence, 26.94 grams of Cl is required to completely react with 35.0 grams of Na.
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