Producing Quicklime (CaO): Volume of CO2 from Decomposition of CaCO3

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SUMMARY

The forum discussion focuses on calculating the volume of carbon dioxide (CO2) produced from the thermal decomposition of calcium carbonate (CaCO3) into quicklime (CaO). The reaction is represented as CaCO3 (s) → CaO (s) + CO2 (g). From 152 grams of CaCO3, the decomposition yields approximately 66.82 grams of CO2, which corresponds to 1.52 moles. Using the ideal gas law (PV=nRT), the calculated volume of CO2 at standard temperature and pressure (STP) is 33.82 liters.

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Homework Statement


Quicklime (CaO) is produced by the thermal decomposition of calcium carbonate
(CaCO3).

Homework Equations


Calculate the volume of CO2 produced at STP from the decomposition
of 152 g of CaCO3 according to the reaction
CaCO3 (s) -----> CaO(s) + CO2(g)


The Attempt at a Solution


isn't 22.4L ?
all i could find is the mass of co2
CaCO3 (s) -----> CaO(s) + CO2(g)
100.08g/mol 44g/mol
152g x
x = 66.82g of CO2
mco2 = 66.82
but then how to find the volume?
 
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Use PV=nRT

You know everything but V. Watch your units so everything comes out in L.
 
P=1atm
T=273,15K
n=1.51mol
R=0.082
V=33.82L
True?
 
Hello people
 
Your multiplication is wrong. Try again.
 
PV = nRT
V=(nRT)/P
V=(1.51*0.082*273.15)/1
V=33.82L
 
That part was right. What did you get for n?
 
what do you mean ? n = 1.51mol
 
66.82g of CO2/44g CO2/mol CO2 = 1.518636364 mol CO2 = 1.52 mol CO2

V=(1.52*0.082*273.15)/1 = ?

Do you need to use significant figures in your answer? If so, R is not 0.082 as well.
 
  • #10
No, they didn't ask to round on the answer.
 

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