Question

For the following reaction, identify the limiting reactant and calculate the grams of product produced when 25.0 grams of the first reactant and 40.0 grams of the second reactant are used: (3 pts)

2SO2 + O2 --> 2SO3

You will have two answers for this question –

Limiting reactant
amount of SO3 produced (grams)

Answer 1

Answer:

The answer to your question is a) limiting reactant SO₂

                                                     b)mass of SO₃ produced 31.25 g

Explanation:

Data

mass of SO₂ = 25 g

mass of O₂ = 40 g

Balanced Reaction

                                     2SO₂   +   O₂   ⇒   2SO₃

Process

1.- Calculate the molecular mass of the reactants

SO₂ = 32 + (16 x 2) = 32 + 32 = 64

O₂ = 16 x 2 =32

2.- Use proportions to determine the limiting reactant

Theoretical proportion    2(SO₂) / O₂ = 2(64) / 32 = 4

Experimental proportion   SO₂ / O₂ = 25 / 40 = 0.625

From these results, we determined that the limiting reactant is SO₂, because the proportion in the experiment is lower.

3.- Amount of SO₃ produced

Mass of SO₃ = 32 + (3 x 16) = 80 g

                        128 g of SO₂   ---------------  2(80) g of SO₃

                        25 g of SO₂    ---------------   x

                         x = (25 x 2 x 80) / 128

                         x = 31.25 g of SO₃