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11111nata11111 [884]
3 years ago
3

A chemist reacted 11.50 grams of sodium metal with an excess amount of chlorine gas. The chemical reaction that occurred is show

n.
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.
Chemistry
1 answer:
Alex_Xolod [135]3 years ago
4 0

Answer:

24.85 g.

Explanation:

  • For the balanced reaction:

<em>2Na + Cl₂ → 2NaCl.</em>

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

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

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.

<u><em>Using stichiometry:</em></u>

∵ 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.

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

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