CaCO3(s) CaO(s) + CO₂(g)
3.25 mol CaCO3 decomposes according to the reaction above. What volume of CO2 gas is produced once the gas cools to STP?

Question

03-08-2022
Chemistry

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CaCO3(s) CaO(s) + CO₂(g)
3.25 mol CaCO3 decomposes according to the reaction above. What volume of CO2 gas is produced once the gas cools to STP?

CaCO3s CaOs COg 325 Mol CaCO3 Decomposes According To The Reaction Above What Volume Of CO2 Gas Is Produced Once The Gas Cools To STP class=

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CintiaSalazar CintiaSalazar · Answer 1 · 2022-08-05T11:33:49-04:00

Taking into account the reaction stoichiometry and the definition of STP, 72.8 L of CO₂ gas is produced once the gas cools to STP.

Reaction stoichiometry

In first place, the balanced reaction is:

CaCO₃ → CaO + CO₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

CaCO₃: 1 mole
CaO: 1 mole
CO₂: 1 mole

STP conditions

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

Moles of CO₂ formed

The following rule of three can be applied: if by reaction stoichiometry 1 mole of CaCO₃ form 1 mole of CO₂, 3.25 moles of CaCO₃ form how many moles of CO₂?

[tex]moles of CO_{2} =\frac{3.25 moles of CaCO_{3}x 1 mole of CO_{2}}{1 moleof CaCO_{3}}[/tex]

moles of CO₂= 3.25 moles

Then, 3.25 moles of CO₂ are formed when 3.25 moles of CaCO₃ decomposes.

Now, you can apply the following rule of three: if by definition of STP conditions 1 mole of CO₂ occupies a volume of 22.4 liters, 3.25 moles occupies how much volume?

[tex]volume= \frac{3.25 molesx22.4 L}{1 mole}[/tex]

volume= 72.8 L

Finally, 72.8 L of CO₂ gas is produced once the gas cools to STP.

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STP:

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Sagot :

Reaction stoichiometry

STP conditions

Moles of CO₂ formed

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