# Question #4dec0

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

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

#"CaCO"_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(->) "CaO"_ ((s)) + "CO"_ (2(g))#

Notice that **for every** **mole** of calcium carbonate that undergoes decomposition, you get **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 **yield**, when **mole** of calcium carbonate--the reaction produces **mole** of calcium oxide.

Since

#1".0 kg" = 10^3color(white)(.)"g"#

you can say that your reaction will produce

#1.0 color(red)(cancel(color(black)("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 **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

#color(darkgreen)(ul(color(black)("% 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.