# Question 373a4

The percent yield of $C a O$ is 58.7%.

#### Explanation:

To determine percent yield, you will need both the actual yield and the theoretical yield. The actual yield is the yield from carrying out the experiment, given in the question. However, you will have to mathematically/stoichiometrically calculate the theoretical yield.

In the question, you are told that the reaction produces 6.81 g of CaO. This is your actual yield.

Now you need the theoretical yield. To do this you will have to perform unit conversion/dimensional analysis using the balanced chemical equation.

For the purposes of a balanced equation (and communication of the general method for these questions), I'll just use:

$C a C {O}_{3} \to C a O + C {O}_{2}$

Now, you know you start with 20.7 g of $C a C {O}_{3}$. Start with this in your unit conversion/dimensional analysis:

$20.7 g$ $C a C {O}_{3}$ --> You'll need to convert this to moles so you can use the balanced chemical equation to get to CaO.

If we use the molar mass of $C a C {O}_{3}$, the grams cancel out nicely, leaving moles:

$20.7 g$ $C a C {O}_{3}$ x $\frac{1 m o l \cdot C a C {O}_{3}}{100.09 g \cdot C a C {O}_{3}}$

Now that you have moles of $C a C {O}_{3}$, you can use the mole ratio from the balanced chemical equation to get to moles of $C a O$.

$C a C {O}_{3}$ and $C a O$ have a 1:1 mole ratio. To cancel out the mol of $C a C {O}_{3}$ you'll have to put it on the bottom of the relationship like so:

$20.7 g$ $C a C {O}_{3}$ x $\frac{1 m o l \cdot C a C {O}_{3}}{100.09 g \cdot C a C {O}_{3}}$ x $\frac{1 m o l \cdot C a O}{1 m o l \cdot C a C {O}_{3}}$

Now that the moles of $C a C {O}_{3}$ have been canceled out, you are left with moles of $C a O$.

Now, you are looking for grams of $C a O$ how can you get to this from where you are? You need a relationship that cancels out moles of $C a O$ and leaves grams of $C a O$. This can come from the molar mass. You'll need moles on the bottom to cancel out the moles of $C a O$. Adding to the previous expression:

$20.7 g$ $C a C {O}_{3}$ x $\frac{1 m o l \cdot C a C {O}_{3}}{100.09 g \cdot C a C {O}_{3}}$ x $\frac{1 m o l \cdot C a O}{1 m o l \cdot C a C {O}_{3}}$ x $\frac{56.08 g \cdot C a O}{1 m o l \cdot C a O}$

Now, if you've done everything correctly, all the units should cancel out, leaving you with grams of $C a O$. Now if you perform all of these calculations you should get:

$11.6 g$ $C a O$

Now to calculate the percentage yield the expression is:

% yield = (actual yield / theoretical yield) * 100%

Substituting the information we have we get:

%*yield = (6.81 g)/(11.6g)(100%)

%*yield = 58.7%#

Hopefully, although long, everything was clear and explained in enough detail! If you have any questions, let me know! :)