At standard temperature and pressure (STP), what is the volume occupied by 12.0 grams of oxygen gas (O_2)? (Atomic mass: O = 15.99 grams/mole)
A
22.4 L
B
8.40 L
C
12.0 L
D
8.37 L
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1
Identify the molar mass of oxygen gas (O_2). Since each oxygen atom has an atomic mass of 15.99 g/mol, calculate the molar mass of O_2 by multiplying by 2: \(\text{Molar mass of } O_2 = 2 \times 15.99 \text{ g/mol}\).
Calculate the number of moles of oxygen gas in 12.0 grams using the formula: \(n = \frac{\text{mass}}{\text{molar mass}} = \frac{12.0 \text{ g}}{\text{molar mass of } O_2}\).
Recall that at standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.4 liters.
Use the relationship between moles and volume at STP to find the volume occupied by the oxygen gas: \(V = n \times 22.4 \text{ L}\).
Substitute the number of moles calculated in step 2 into the equation in step 4 to find the volume of oxygen gas at STP.