Answer:
18.0g of CO₂ are present in 9.16L
Explanation:
To solve this question we must use:
PV = nRT
In order to find the moles of the gas and with its molar mass (44.01g/mol for CO₂) we can find the mass of the gas
Assuming STP conditions:
P = 1atm at STP
V = 9.16L
n are the moles of CO₂
R = 0.082atmL/molK
T = 273.15K
Replacing:
PV / RT = n
1atm*9.16L / 0.082atmL/molK*273.15K = n
0.409 moles = Moles of CO₂
The mass is:
0.409 moles CO₂ * (44.01g / mol) =
18.0g of CO₂ are present in 9.16L
