# Question #ef8fc

##### 2 Answers

#### Explanation:

Like always, start with a balanced chemical equation for this **decomposition reaction**

#"CaCO"_text(3(s]) stackrel(color(red)(Delta)color(white)(aa))(->) "CaO"_text((s]) + "CO"_text(2(g])# #uarr#

Notice that you have **mole ratios** across the board. **One mole** of calcium carbonate will decompose to form **one mole** of calcium oxide and **one mole** of carbon dioxide.

Keep this in mind.

Now, since you didn't provide information about the *temperature* and *pressure* at which the oxygen gas is being collected, I"ll assume that you're at **STP**, **Standard Pressure and Temperature**.

Under STP condittions, which are defined as a pressure of **one mole** of any ideal gas is known to occupy **molar volume of a gas at STP**.

This means that you can use the volume of carbon dioxide to figure out how many moles of oxygen were produced by the reaction

#6.72 color(red)(cancel(color(black)("L"))) * overbrace("1 mole O"_2/(22.7color(red)(cancel(color(black)("L")))))^(color(purple)("molar volume of a gas")) = "0.2960 moles CO"_2#

Now go back to the mole ratio that exists between calcium carbonate and carbon dioxide. If the reaction produced

Use calcium carbonate's **molar mass** to find how many grams of calcium carbonate would contain this many moles

#0.2960 color(red)(cancel(color(black)("moles CaCO"_3))) * "100.1 g"/(1color(red)(cancel(color(black)("mole CaCO"_3)))) = "29.63 g"#

Now, you know that your initial sample contained **impurities**. This is equivalent to saying that it contained

If that is the case, then the initial mass of the sample was

#29.63 color(red)(cancel(color(black)("g CaCO"_3))) * overbrace("100 g sample"/(80color(red)(cancel(color(black)("g CaCO"_3)))))^(color(purple)(=80%"CaCO"_3)) = color(green)("37 g")#

I'll leave the answer rounded to two **sig figs**.

**SIDE NOTE** *Make sure to use the value of the molar volume of a gas at STP given to you. Usually, that value is* *so if that's what you have, simply redo the calculations by replacing* *with your value*.

If the volume of CO2 ie 6.72L is in STP, then calculation may be made as follows

#### Explanation:

According to the reaction CaCO3 (s)------> CaO(s) + CO2(g)

1 mole or 22.4 L CO2 at STP is obtained on thermal decomposition of 100 g of CaCO3,

So 6.72L CO2 can be obtained from 30 g CaCO3. As the sample contains 20% impurities,it can be said that 80 g CaCO3 (which will only produce CO2) is present in 100 g of sample.Hence the initial mass of the sample should be (100X30)/80=37.5 g