Achemist reacted 11.50 grams of sodium metal with an excess amount of chlorine gas. the chemical reaction that occurred is shown.

na + cl2 → nacl

if the percentage yield of the reaction is 85%, what is the actual yield? show your work, including the use of stoichiometric calculations and conversion factors.

24.85 g.

Explanation:

2Na + Cl₂ → 2NaCl.

It is clear that 2.0 moles of Na metal react with 1.0 mol of Cl₂ to produce 2.0 moles of NaCl.

∵ Yield% = [actual yield/theoretical yield] x 100.

To get the theoretical yield:

We need to calculate the no. of moles of 11.5 g Na metal:

n = mass/atomic mass = (11.5 g)/(22.989 g/mol) = 0.5 mol.

Using stichiometry:

∵ 2.0 moles of Na metal produces 2.0 moles of NaCl.

∴ 0.5 mol of Na will produce 0.5 mol of NaCl.

So, the no. of grams of NaCl (theoretical yield) = no. of moles x molar mass = (0.5 mol)(58.44 g/mol) = 29.23 g.

∵ Yield% = [actual yield/theoretical yield] x 100.

∴ actual yield = (Yield%)(theoretical yield)/100 = (85%)(29.23 g)/100 = 24.85 g.

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