Consider the following data: s(s) + o2(g) → so2(g)    δhrxn = -297 kj 2so3(g) → 2so2(g) + o2(g)  δhrxn = 198 kj use hess's law to determine ^hrxn for this reaction: 2s(s) + 3o2(g) > 2so3(g) ^hrxn = ?

-792 KJ.

Explanation:

To get ΔH of the reaction: 2S + 3O₂ → 2SO₃ The sum of the two mentioned equations should give the result of the reaction we need to calculate its ΔH.We multiply the first equation by 2 and also its ΔHrxn:

2S + 2O₂ → 2SO₂             ΔHrxn = -594 KJ

and reverse the second reaction and so, multiply the ΔHrxn by -1:

2SO₂ + O₂ → 2SO₃          ΔHrxn = -198 KJ

By summing the two reactions, we get the reaction that we need to calculate its ΔHrxn: 2S + 3O₂ → 2SO₃ Now, we can apply Hess's law says that ΔH is a state function that do not depend on the path of the reaction but depends on the initial and final state.ΔHrxn = ΔH of the first reaction + ΔH of the second reaction ΔHrxn = (-594 KJ) + (-198 KJ) = -792 KJ.

4777.09grams

Explanation:

To find the mass of oxygen, we first need to find the mass of potassium sulfate (K2SO4).

Since there are 4.50 x 10^25 formula units of potassium sulfate, we can find the number of moles in K2SO4 by dividing by Avagadros number (6.02 × 10^23 units). That is;

number of moles of K2SO4 (n) = 4.50 x 10^25 ÷ 6.02 × 10^23

= 0.747 × 10^ (25-23)

= 0.747 × 10^2

= 7.47 × 10^1 moles

Mass in grams of K2SO4 can be calculated thus: molar mass of K2SO4 × moles

= 174.252 g/mol × 7.47 × 10^1 moles

= 13016.62grams.

Since mass of oxygen in 1 mol of K2SO4 = O4 = 63.996 g/mol

We find the percentage by mass of oxygen in K2SO4 as follows:

= 63.996/174.252 × 100

= 0.367 × 100

= 36.7% by mass of oxygen.

This means that in 1gram of K2SO4, there are 0.367gram of Oxygen. Hence, in 13016.62grams (4.50 x 10^25 units) of K2SO4, there will be;

0.367 × 13016.62

= 4777.09grams of oxygen.