CHEMISTRY PROBLEM

Question

50.0 mL of 0.10 M sodium chloride (NaCl) solution is mixed with 25.0 mL of 0.25 M silver nitrate (AgNO₃) solution. After filtering and drying the product. 0.20 g of silver chloride was produced. What is the percent yield from this reaction?

a	28%
b	47%
c	69%
d	94%

Answer 1

First, write a balanced equation:

NaCl(aq) + AgNO₃(aq) ==> AgCl(s) + NaNO₃(aq)

Next, find the moles of each reactant:

moles NaCl = 0.050 L x 0.10 mol/L = 0.0050 moles NaCl

moles AgNO₃ = 0.0250 L x 0.25 mol/L = 0.00625 moles AgNO₃

Now, determine which reactant is limiting:

Since the mole ratio of NaCl : AgNO₃ is 1 : 1, the NaCl is limiting because is is present in smallest amount.

Determine moles of AgCl formed:

0.0050 moles NaCl x 1 mole AgCl/ mol NaCl = 0.005 moles AgCl

Determine mass of AgCl:

0.005 moles AgCl x 143.3 g/mol = 0.7165 g AgCl

Finally, determine percent yield:

actual yield/theoretical yield (x100%) = 0.20 g/0.7165 g (x100%) = 27.9% = 28% (Answer a)