If a person inhales 91.5 g of O2 in an hour, what volume (in L) does this amount occupy at 1.09 atm and 36.8 °C? Be sure your answer has the correct number

Question

of significant figures.
Note: Reference the Fundamental constants table for additional information.

Answer 1

To solve this problem, we need to use the ideal gas law formula:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)

First, we need to calculate the number of moles of O2 in 91.5g:

1. Calculate the molar mass of O2:
O = 16.00 g/mol
2 * O = 2 * 16.00 = 32.00 g/mol

2. Calculate number of moles:
91.5g / 32.00 g/mol = 2.859 moles

Next, we convert the temperature to Kelvin:

- 36.8 °C + 273.15 = 309.95 K

Now we can plug in the values into the ideal gas law formula:

(1.09 atm) * V = (2.859 mol) * (0.0821 L.atm/mol.K) * (309.95 K)

V = (2.859 * 0.0821 * 309.95) / 1.09
V = 67.27 L

Therefore, 91.5 g of O2 at 1.09 atm and 36.8 °C occupies 67.27 liters of volume.