Question
How many grams of CO, gas occupy 40.1 L at STP? Be sure your answer has the correct number of significant figures.
Note: Reference the Fundamental Constants table for additional information.
Note: Reference the Fundamental Constants table for additional information.
Answers
GPT 3.5
Using the ideal gas law, we can calculate the number of moles of CO gas using the formula:
PV = nRT
Where:
P = pressure (STP = 1 atm)
V = volume (40.1 L)
n = number of moles
R = ideal gas constant (0.08206 atm·L/mol·K)
T = temperature (STP = 273.15 K)
Plugging in the values:
(1 atm) (40.1 L) = n (0.08206 atm·L/mol·K) (273.15 K)
40.1 = 22.414 n
n = 40.1 / 22.414
n ≈ 1.79 moles
Next, we can convert the moles of CO gas to grams using the molar mass of CO (28.01 g/mol for C + 16.00 g/mol for O = 44.01 g/mol for CO):
1.79 moles x 44.01 g/mol = 78.80 g
Therefore, 78.80 grams of CO gas occupy 40.1 L at STP. The answer has three significant figures.
PV = nRT
Where:
P = pressure (STP = 1 atm)
V = volume (40.1 L)
n = number of moles
R = ideal gas constant (0.08206 atm·L/mol·K)
T = temperature (STP = 273.15 K)
Plugging in the values:
(1 atm) (40.1 L) = n (0.08206 atm·L/mol·K) (273.15 K)
40.1 = 22.414 n
n = 40.1 / 22.414
n ≈ 1.79 moles
Next, we can convert the moles of CO gas to grams using the molar mass of CO (28.01 g/mol for C + 16.00 g/mol for O = 44.01 g/mol for CO):
1.79 moles x 44.01 g/mol = 78.80 g
Therefore, 78.80 grams of CO gas occupy 40.1 L at STP. The answer has three significant figures.