Answer
0.9154 mol CO₂
Procedure
To solve this question you will need to use the ideal gas law
PV=nRT
Data
P=4.55 atm
V= 3.5 L
T= 212 K
R=0.082057 L⋅atm⋅K⁻¹⋅mol⁻¹
Solve for n
[tex]n=\frac{PV}{RT}=\frac{4.55\text{ atm }3.5\text{L mol }\degree\text{K}}{0.082057\text{ L.atm }212\degree\text{K}}=0.9154\text{ mol }[/tex]