What volume would 5.5 gram of CO2 occupy at STP?
To calculate the volume of 5.5 grams of CO2 at STP (Standard Temperature and Pressure), we need to use the ideal gas law.
The ideal gas law is PV = nRT, where:
- P is the pressure in atmospheres
- V is the volume in liters
- n is the number of moles of gas
- R is the ideal gas constant (0.0821 L·atm/mol·K)
- T is the temperature in Kelvin
At STP, the pressure is 1 atm and the temperature is 273.15 K.
First, we need to calculate the number of moles of CO2:
molar mass of CO2 = 12.01 g/mol + 2 * 16.00 g/mol = 44.01 g/mol
moles of CO2 = mass of CO2 / molar mass of CO2
= 5.5 g / 44.01 g/mol
≈ 0.125 moles
Now, we can use the ideal gas law to calculate the volume:
PV = nRT
V = nRT / P
= (0.125 moles) * (0.0821 L·atm/mol·K) * (273.15 K) / (1 atm)
≈ 3.49 liters
Therefore, 5.5 grams of CO2 would occupy approximately 3.49 liters at STP.