32 g of sulfur will react with 48 g of oxygen to produce 80 g of sulfur trioxide. If 32 g of sulfur and 100 g of oxygen are placed into a sealed container and allowed to react, what is the mass of the material in the container after the reaction is completed?(A) 100 g(B) 132 g(C) 80 g(D) 32 g

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)