Final answer:
To calculate the grams of SO₂ gas at STP in a 5.9 L container, the volume is divided by the molar volume at STP to find moles, which is then multiplied by the molar mass of SO₂ to find the mass, resulting in approximately 16.87 grams of SO₂.
Step-by-step explanation:
Calculating Grams of SO₂ Gas at STP in a 5.9 L Container
To calculate the grams of SO₂ gas present at STP (standard temperature and pressure) in a 5.9 L container, we first need to use the ideal gas law equation. At STP, one mole of any gas occupies 22.4 liters. First, we find the number of moles (n) of SO₂ using the volume of the container and the molar volume of a gas at STP:
n = Volume / Molar Volume at STP = 5.9 L / 22.4 L/mol
Now we calculate the moles of SO₂:
n = 5.9 L / 22.4 L/mol ≈ 0.2634 mol
Next, we use the molar mass of SO₂ to convert moles to grams:
Mass = n × Molar Mass = 0.2634 mol × 64.07 g/mol ≈ 16.87 grams of SO₂
Therefore, there are approximately 16.87 grams of SO₂ gas in a 5.9 L container at STP.