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

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Discussion Overview

The discussion revolves around calculating the volume of carbon dioxide (CO2) produced from the thermal decomposition of calcium carbonate (CaCO3) to quicklime (CaO). It includes aspects of stoichiometry, gas laws, and unit conversions, framed as a homework problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that the volume of CO2 produced is 22.4 L based on their calculations.
  • Another participant advises using the ideal gas law (PV=nRT) to find the volume, indicating that the volume is the unknown variable.
  • Participants provide values for pressure (P=1 atm), temperature (T=273.15 K), and the gas constant (R=0.082), leading to a calculation of 33.82 L for the volume of CO2.
  • There is a correction regarding the calculation of moles (n), with one participant asserting that n equals 1.51 mol based on their computations.
  • Another participant recalculates n as approximately 1.52 mol CO2 and questions the need for significant figures in the final answer, suggesting that R should not be rounded to 0.082.
  • One participant states that rounding was not required for the answer.

Areas of Agreement / Disagreement

Participants generally agree on the use of the ideal gas law for the calculation, but there is some disagreement regarding the correct value of n and the implications of significant figures in the calculations.

Contextual Notes

The discussion does not resolve the exact volume of CO2 produced due to differing calculations and assumptions about significant figures and rounding.

chawki
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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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