Answer:
Since the container is consealed, and O2 will no be completely consumed, the total mass of material in the container will be 80 grams SO3+ 52 grams O2 = 132 grams (option B)
Explanation:
Step 1: Data given
Mass of sulfur = 32.00 grams
Mass of oxygen = 48.00 grams
Molar mass of sulfur = 32.07 g/mol
Molar mass of oxygen = 32 g/mol
Molar mass of SO3 = 80.07 g/mol
Step 2: The balanced equation
2S + 3O2 → 2SO3
Step 3: Calculate moles S
Moles S = Mass S / molar mass S
Moles S = 32.0 grams / 32.07 g/mol
Moles S = 0.998 moles
Step 4: Calculate moles O2
Moles O2 = 100.0 grams / 32.0 g/mol
Moles O2 = 3.125 moles
Step 5: Calculate the limiting reactant
For 2 moles S we need 3 moles O2 to produce 2 moles SO3
S is the limiting reactant. It will completely be consumed (0.998 moles)
O2 is in excess, there will be consumed 3/2 * 0.998 = 1.497 moles
There will remain 3.125- 1.497 = 1.628 moles O2
This is 1.628 moles * 32 g/mol = 52.1 grams
Step 6: Calculate moles SO3
For 2 moles S we need 3 moles O2 to produce 2 moles SO3
For 0.998 moles S there will react 0.998 moles SO3
Step 6: Calculate mass SO3
Mass SO3 = moles SO3 * molar mass SO3
Mass SO3 = 0.998 moles * 80.07 g/mol
Mass SO3 = 79.9 grams ≈ 80 grams
There will be produced 80 grams of SO3
Since the container is consealed, and O2 will no be completely consumed, the total mass of material in the container will be 80 grams SO3+ 52 grams O2 = 132 grams (option B)
