Final answer:
To calculate the mass of oxygen in the mixture, the ideal gas law is used to determine the total moles of gas in the flask, from which the known moles of methane are subtracted to obtain the moles of oxygen, which are then converted to grams.
Step-by-step explanation:
The number of grams of oxygen in the mixture is 36.67 g.
To solve for the mass of oxygen in the mixture, we first use the Ideal Gas Law, PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is temperature in Kelvin. However, instead of directly calculating the moles of oxygen, we first find the moles of methane (CH4) since its mass is given. The molar mass of methane is about 16.04 g/mol, so 7.42 g of methane equals approximately 0.462 moles. Using the Ideal Gas Law rearranged to n = PV/(RT) and plugging in the values for methane, we can find the total moles of the gas mixture in the flask.
Once we know the total moles of the gas mixture and the moles of methane, we subtract to find the moles of oxygen. Finally, we convert the moles of oxygen to grams using the molar mass of oxygen (32.00 g/mol).
PV = (nCH4 + nO2)RT
For methane, nCH4 = 7.42 g / 16.04 g/mol = 0.462 moles
R (ideal gas constant) = 0.0821 L*atm/mol*K
The temperature must be in Kelvin: T = 85 °C + 273.15 = 358.15 K
Now, calculate the total moles of gas (ntotal) in the flask: ntotal = PV/RT = (2.36 atm * 8.43 L) / (0.0821 L*atm/mol*K * 358.15 K) = 0.711 moles.
The moles of oxygen gas (nO2) in the flask will then be ntotal - nCH4 = 0.711 moles - 0.462 moles = 0.249 moles.
Finally, convert nO2 to grams: massO2 = nO2 * molar mass of O2 = 0.249 moles * 32.00 g/mol = 36.67 g.