# When 60g of calcium carbonate is heated, 16.8g of Calcium oxide are formed. What is the percentage yield? (multiple choice)

## When calcium carbonate is heated it decomposes according to the equation: ${\text{CaCO"_3 -> "CaO" + "CO}}_{2}$ A. 16.8% B. 25% C. 50% D. 60% The answer is C. 50%, why?

Feb 17, 2016

"% yield" = 50%

#### Explanation:

Start by taking a second to make sure that you understand what the problem wants you to determine.

A reaction's percent yield will essentially tell you one thing - how efficient said reaction is.

In this context, the more efficient a reaction, the closer it will be to theoretical production, which is what we calculate assuming that absolutely every molecule of the reactants will actually react to form the desired product.

The theoretical yield of a reaction will thus correspond to a 100% yield. The actual yield, which is what you get in practice, will correspond to the reaction's percent yield.

So, the problem wants you to figure out how efficient this reaction is at producing calcium oxide, $\text{CaO}$, and ${\text{CO}}_{2}$, starting from the given mass of calcium carbonate, ${\text{CaCO}}_{3}$.

${\text{CaCO"_text(3(s]) stackrel(color(red)(Delta)color(white)(aa))(->) "CaO"_text((s]) + "CO}}_{\textrm{2 \left(g\right]}}$ $\uparrow$

So, what would you theoretical yield be?

Well, in theory, every mole of calcium carbonate that undergoes decomposition should produce one mole of calcium oxide.

Take a look at the 8molar masses* of the two compounds

${\text{For CaCO"_3: " "M_M = "100.09 g mol}}^{- 1}$

${\text{For CaO: " M_M = "56.08 g mol}}^{- 1}$

So, if one mole of calcium carbonate should produce one mole of calcium oxide, it follows that $\text{100.09 g}$ of calcium carbonate should produce $\text{56.08 g}$ of calcium oxide.

This means that $\text{60 g}$ of calcium carbonate should produce

60 color(red)(cancel(color(black)("g CaCO"_3))) * "56.08 g CaO"/(100.09color(red)(cancel(color(black)("g CaCO"_3)))) = "33.65 g CaO"

You know that your reaction actually produced $\text{16.8 g}$ of calcium oxide. This means that its percent yield was

16.8color(red)(cancel(color(black)("g CaO"))) * overbrace("100% yield"/(33.65color(red)(cancel(color(black)("g CaO")))))^(color(purple)("theoretical yield")) = 49.9% ~~ color(green)("50%")

This is exactly what the formula for percent yield gets you

$\textcolor{b l u e}{\text{% yield" = "what you actually get"/"what you should theoretically get} \times 100}$