Volume of CO2 produced at STP from decomposing 152 g of CaCO3

[chawki]

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
m_co2 = 66.82
but then how to find the volume?

 

[chemisttree]

Use PV=nRT

You know everything but V. Watch your units so everything comes out in L.

 

[chawki]

P=1atm
T=273,15K
n=1.51mol
R=0.082
V=33.82L
True?

 

[chawki]

Hello people

 

[chemisttree]

Your multiplication is wrong. Try again.

 

[chawki]

PV = nRT
V=(nRT)/P
V=(1.51*0.082*273.15)/1
V=33.82L

 

[chemisttree]

That part was right. What did you get for n?

 

[chawki]

what do you mean ? n = 1.51mol

 

[chemisttree]

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.

 

[chawki]

No, they didn't ask to round on the answer.