If 86.33 grams of sodium reacts with an excess of chlorine, what amount of Salt (sodium chloride) will be produced? Sodium + chlorine ® sodium chloride Na + Cl2 ® NaCl

Md Ariful Islam

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Solution:

Sodium + chlorine → sodium chloride  

The balanced chemical equation:

2Na + Cl2 → 2NaCl

According to the equation above: n(Na) = n(NaCl)

Moles of Na = Mass of Na / Molar mass of Na

The molar mass of Na is 22.9898 g mol-1.

Hence,

n(Na) = 86.33 g / 22.9898 g mol-1 = 3.755 mol

n(NaCl) = n(Na) = 3.755 mol

Moles of NaCl = Mass of NaCl / Molar mass of NaCl

Mass of NaCl = Moles of NaCl × Molar mass of NaCl

The molar mass of NaCl is 58.44 g mol-1.

Hence,

m(NaCl) = 3.755 mol × 58.44 g mol-1 = 219.44 g

m(NaCl) = 219.44 g

OR:

(86.33 g Na)×(1 mol Na/22.9898 g Na)×(2 mol NaCl/2 mol Na)×(58.44 g NaCl/1 mol NaCl) = 219.45 g

Answer: 219.44 grams (or 3.755 mol) of sodium chloride (NaCl) will be produced.

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