Question

For the following reaction, 25.2 grams of sulfur dioxide are allowed to react with 5.36 grams of oxygen gas. sulfur dioxide (g) oxygen (g) sulfur trioxide (g) What is the maximum amount of sulfur trioxide that can be formed

Answer 1

Answer:

Maximum amount of sulfur trioxide that can be formed = 26.822 g

Explanation:

The balanced chemical equation for the reaction

2SO₂ + O₂ -----> 2SO₃

25.2 grams of sulfur dioxide are allowed to react with 5.36 grams of oxygen gas. What is the maximum amount of sulfur trioxide that can be formed?

It is the limiting reagent (the reactant in the stoichiometric lesser amount) that determines how much product is formed or how much of the other reactant is formed.

So, we convert the masses of reactants present into number of moles to get a clearer picture.

(Number of moles) = (mass)/(molar mass)

For sulfur dioxide,

Mass present = 25.2 g

Molar mass = 64.066 g/mol

(Number of moles present) = (25.2/64.066)

(Number of moles present) = 0.39 moles

For Oxygen gas,

Mass present = 5.36 g

Molar mass = 32.0 g/mol

(Number of moles present) = (5.36/32)

(Number of moles present) = 0.1675 moles

But from the stoichiometric balance,

2SO₂ + O₂ -----> 2SO₃

2 moles of Sulfur dioxide reacts with 1 mole of Oxygen gas

If Sulfur dioxide was the limiting reagent,

0.39 moles would react with (0.39×1/2) moles of Oxygen gas; 0.195 moles of Oxygen gas.

This is more than the total amount of Oxygen gas present at the start of the reaction, hence, Sulfur dioxide cannot be the limiting reagent.

Oxygen gas as limiting reagent,

1 mole of Oxygen gas reacts with 2 moles of Sulfur dioxide,

0.1675 moles of Oxygen gas would react with (0.1675×2/1) of Sulfur dioxide; 0.335 moles of Sulfur dioxide.

This indicates that oxygen is truly the limiting reagent and Sulfur dioxide is the reagent that is present in excess.

So, now, we calculate the amount of Sulfur trioxide that can be obtained from this reaction setup (assuming a 100% conversion and the maximum amount of Sulfur dioxide formed)

2SO₂ + O₂ -----> 2SO₃

1 mole of Oxygen gas gives 2 moles of Sulfur trioxide,

0.1675 moles of Oxygen gas will give (0.1675×2/1) moles of Sulfur trioxide; 0.335 moles of Sulfur trioxide.

We then convert this to mass.

(Mass) = (number of moles) × (molar mass)

Molar mass of SO₃ = 80.066 g/mol

(Mass of SO₃ produced) = 0.335 × 80.066

(Mass of SO₃ produced) = 26.822 g.

Maximum amount of sulfur trioxide that can be formed = 26.822 g

Hope this helps!!

Answer 2

Explanation:

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