# Question 4dec0

Sep 16, 2017

Here's what I got.

#### Explanation:

Start by writing the balanced chemical equation that describes this decomposition reaction

${\text{CaCO"_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(->) "CaO"_ ((s)) + "CO}}_{2 \left(g\right)}$

Notice that for every $1$ mole of calcium carbonate that undergoes decomposition, you get $1$ mole of calcium oxide.

You can use the molar masses of the two compounds

• M_ ("M CaCO"_ 3) = "100.09 g mol"^(-1)
• M_ ("M CaO") = "56.08 g mol"^(-1)

to say that at 100% yield, when $\text{100.09 g}$ of calcium carbonate undergo decomposition--the equivalent of $1$ mole of calcium carbonate--the reaction produces $\text{56.08 g}$ of calcium oxide--the equivalent of $1$ mole of calcium oxide.

Since

$1 \text{.0 kg" = 10^3color(white)(.)"g}$

you can say that your reaction will produce

$1.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{kg CaCO"_3))) * (10^3color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * "56.08 g CaO"/(100.09color(red)(cancel(color(black)("g CaCO"_3)))) = color(darkgreen)(ul(color(black)("560 g CaO}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the mass of calcium carbonate that undergoes decomposition.

Now, you know that the reaction produced

0.5 color(red)(cancel(color(black)("kg"))) * (10^3 color(white)(.)"g")/(1color(red)(cancel(color(black)("kg")))) = "500 g"

of calcium oxide. In order to find the percent yield of the reaction, you need to figure out how many grams of calcium oxide are produced for every $\text{100 g}$ of calcium oxide that could theoretically be produced.

100 color(red)(cancel(color(black)("g in theory"))) * "500 g CaO produced"/(560 color(red)(cancel(color(black)("g in theory")))) = "90 g CaO"#

You can thus say that the reaction's percent yield is equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{% yield = 90%}}}}$

The answer is rounded to one significant figure, the number of sig figs you have for the mass of calcium oxide produced by the reaction.