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# Explanation of the way to use the ideal gas law with solids

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What volume of carbon dioxide at STP is produced by the complete decomposition of 24 g of CaCO3--2.69 dm3, 4.64 dm3,5.37 dm3, 7.68 dm3

To solve the question we write down the equation of the reaction:

CaCO3 -----> CO2 + CaO

This equation is balanced out. The number of atoms of every element is the same on the left and right side.
What we can see is that 1 mole CaCO3 gives 1 mole CO2.

To find out the correct volume of CO2 we have to do the following steps:

1) Transforming the given mass [g] of CaCO3 into mole gives us the number of moles of CO2.

2) We need to calculate the molar mass of both CaCO3 and CO2. To do that we need the atomic mass of every atom of CaCO3 and CO2.

in a table of elements we find:

Ca : 40.078 u
C : 12.011 u
O : 15.999 u

1 mole of CaCO3 has the molar mass : 40.078 + 12.011 + (3 * 15.999) = 100.086 g

In the same way we get the molar mass of CO2 ( 40.009 g )

Now, we can see in case of CaCO3:

100.086 g = 1 mole
24 g = 100.086 / 24 = 0.239 mole

Based on our equation we can see that 1 mole CaCO3 gives 1 mole of CO2.
That means that 0.239 mole CaCO3 give 0.239 mole CO2

2) Using the relation ' 1 mole of a gaseous compound has a volume of 22.41 dm3 ' gives us the volume of the formed CO2 .

We find:

1 mole = 22.41 dm3
0.239 mole = 0.239 * 22.41 = 5.37 dm3

=> you have calculated the correct value :)
Thanks for using BrainMass.com help service

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https://brainmass.com/chemistry/gas-laws/explanation-of-the-way-to-use-the-ideal-gas-law-with-solids-44087

#### Solution Preview

To solve the question we write down the equation of the reaction:

CaCO3 -----> CO2 + CaO

This equation is balanced out. The number of atoms of every element is the same on the left and right side.
What we can see is that 1 mole CaCO3 gives 1 mole CO2.

To find out the correct volume of CO2 we have to do ...

#### Solution Summary

How to find out the correct chemical equation, the molecular masses and how to calculate the amount of product using the ideal gas law. The general procedure described also is useful for similar problems. The solution is original and a step-by-step one.

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