Question

When 15 grams of copper (II) chloride (CuCl2) reacts with 20 grams of sodium nitrate (NaNO3), 11.3 grams of sodium chloride (NaCl) is obtained. What is the percent yield of this reaction?

Answer 1

balanced equation:

CuCl2 + 2NaNO3 → Cu(NO3)2 + 2NaCl

first, you'll need to figure out whether CuCl2 or NaNO3 is the limiting reactant. the limiting reactant will produce the lower amount of product.

so, you can convert both reactants to product, using stoichiometry, and see which one is lower.

setting up the stoich:

\[\frac{ 20g~NaNO3 }{ }\frac{ 1~mol~Na }{ 84.995g~NaNO3 }\frac{ 2~mol~NaCl }{ 2~mol~NaNO3 }\]

\[\frac{ 15g~CuCl2 }{ }\frac{ 1~mol~CuCl2 }{ 134.45g~CuCl2}\frac{ 2~mol~NaCl }{ 1~mol~CuCl2 }\]

Answer 2

from there, once you have the limiting reactant, you can keep going where you left off with the stoich. we are now in moles NaCl, so you can simply multiply by the molar mass of NaCl to get the *theoretical* yield

finally, to get the % yield, calculate (actual yield / theoretical yield) * 100%. don't forget to round to the appropriate # of sig figs at the end.