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In a salt synthesis experiment, 13.1 g of sodium reacts with 94 g of chlorine gas, yielding 22.5 g of sodium chloride. 2Na(s) + Cl2(g) →2NaCl(s) What is the percent yield for this experiment?

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Answer:

67.58%  

Explanation:

First we convert the masses of the reactants into moles, using their respective molar masses:

  • Na ⇒ 13.1 g ÷ 23 g/mol = 0.570 mol
  • Cl₂ ⇒ 94 g ÷ 70.9 g/mol = 1.32 mol

By looking at the reaction equation, we know that 1.32 moles of Cl₂ would react completely with (2 * 1.32) 2.64 moles of Na. There are not as many Na moles, so Na is the limiting reactant.

Now we calculate how many NaCl moles would have been produced if all the limiting reactant was consumed:

  • 0.570 mol Na * [tex]\frac{2molNaCl}{2molNa}[/tex]= 0.570 mol NaCl

We convert those moles to grams:

  • 0.570 mol * 58.44 g/mol = 33.31 g NaCl

Finally we calculate the percent yield:

  • 22.5 g / 33.3 g * 100% = 67.58%  

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