Question

Calculate the grams of SO3 that you would make in the following reaction when 20g of SO2 and 40 g of O2 are used in the reaction. 2SO2(g) + O2(g) ---> 2SO3(g)

Answer 1

25.01g

Step-by-step explanation:

2SO2(g) + O2(g) ---> 2SO3(g)

SO2 = 20g

O2 = 40g

1. To determine how many grams of SO3 would form, we need to find the

limiting reaction. (The reactant that will be consumed completely first) To do so:

SO2: 20g x Molar Mass of SO2

20g x ( 1 mol / 64.066g)

SO2 = 0.312 mols <-- This is the limiting reagent because it has the

smallest amount of moles compared to O2

O2: 40g x Molar Mass of O2

40g x ( 1 mol / 32.0g )

O2 = 1.25 mols

2. Now using the mole ratio of the limiting reagent to the product can tell how much SO3 is formed:

2:2 mole ratio ( Was determined by the compound coefficients.)

Since the mole ratio is 1:1, we do not need to double or half the amount of moles of the limiting reagent to determine the product formed.

3. Calculating the SO3 formed:

(0.312 g/mol) x (Molar mass of SO3) = # grams of SO3

(0.312 g/mol) x (80.06 g/mol) = # grams of SO3

25.01g = # grams of SO3