Question

In a reaction, 20.50g of NaOH is reacted with excess Cu(NO3)2.

2 NaOH(aq) + Cu(NO3)2(aq) → Cu(OH)2(s) + 2NaNO3(aq)

a. What is the theoretical yield of Cu(OH)2?

b. If the actual yield of copper(II) hydroxide was determined to be 75.0g, what is the
percent yield?

Answer 1

a. To determine the theoretical yield of Cu(OH)2, we need to first calculate the limiting reactant. The balanced chemical equation shows that 2 moles of NaOH reacts with 1 mole of Cu(NO3)2 to produce 1 mole of Cu(OH)2. We can use this information to calculate the number of moles of NaOH:

20.50 g NaOH x (1 mole NaOH / 40.00 g NaOH) = 0.5125 moles NaOH

Using the stoichiometry of the balanced equation, we can see that this amount of NaOH will react with:

0.5125 moles NaOH x (1 mole Cu(NO3)2 / 2 moles NaOH) = 0.2563 moles Cu(NO3)2

Therefore, 0.2563 moles Cu(NO3)2 will produce:

0.2563 moles Cu(OH)2 x (97.56 g Cu(OH)2 / 1 mole Cu(OH)2) = 25.00 g Cu(OH)2

So, the theoretical yield of Cu(OH)2 is 25.00 g.

b. The percent yield is calculated as the actual yield divided by the theoretical yield, multiplied by 100%:

Percent yield = (actual yield / theoretical yield) x 100%

Percent yield = (75.0 g / 25.00 g) x 100% = 300%

The percent yield is greater than 100%, which indicates that there may have been errors in the experiment that led to a higher than expected yield.