A 25.0 L sample of oxygen gas (O_2) is collected at 5.20 atm and 28.0 °C. Using the ideal gas law, how many grams of O_2 are present in the sample?

• A. 170 g
• B. 104 g
• C. 36.2 g
• D. 58.7 g

Answer: (A)

Verified step by step guidance

1. Identify the known variables from the problem: volume \(V = 25.0\ \text{L}\), pressure \(P = 5.20\ \text{atm}\), and temperature \(T = 28.0\ ^\circ\text{C}\). Convert the temperature to Kelvin using the formula \(T(K) = T(^\circ\text{C}) + 273.15\).
2. Write down the ideal gas law equation: \(P \times V = n \times R \times T\), where \(n\) is the number of moles of gas, and \(R\) is the ideal gas constant. Use \(R = 0.0821\ \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K}\) for consistency with the given units.
3. Rearrange the ideal gas law to solve for the number of moles \(n\): \(n = \frac{P \times V}{R \times T}\).
4. Calculate the number of moles \(n\) by substituting the known values of \(P\), \(V\), \(R\), and \(T\) (in Kelvin) into the equation.
5. Convert moles of \(O_2\) to grams by multiplying the number of moles \(n\) by the molar mass of oxygen gas. The molar mass of \(O_2\) is approximately \$32.00\ \text{g/mol}\(. Use the formula: \)\text{mass} = n \times 32.00\ \text{g/mol}$.