What is the error? NH4OH (aq) + KOH (aq) --> KOH (aq) + NH4OH (aq)

Chemistry

1 answer:

dexar [7]3 years ago

6 0

Answer:

There is not a chemical reaction. This is just the dissolution of two ionic compounds that remain dissolved in solution, without reacting.

Explanation:

The issue here is that none chemical reaction is happening.

The two compounds mixed are ionic bases: both of them dissociate in water producing OH⁻ ions:

NH₄OH (aq) → NH₄⁺(aq) + OH⁻(aq)

KOH (aq) → K⁺(aq) + OH⁻(aq)

Then, when you write the total ionic equation you have:

NH₄⁺(aq) + OH⁻(aq) + K⁺(aq) + OH⁻(aq) →

NH₄⁺(aq) + OH⁻(aq) + K⁺(aq) + OH⁻(aq)

Look at the species on the reactant side and the product side: the same ions are found on both sides.

That means that all of them are spectator ions: they are just dissolved, and remain in water (aqueous solution) without reacting. None new different compound is formed.

The net ionic equation would not have any species because the spectator ions get canceled.

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The formation of SO3 from SO2 and O2 is an intermediate step in the manufacture of sulfuric acid, and it is also responsible for

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Answer:

118.22 atm

Explanation:

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

KP = 0.13 = $\frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}$

Where p(SO₃) is the partial pressure of SO₃, p(SO₂) is the partial pressure of SO₂ and p(O₂) is the partial pressure of O₂.

With 2.00 mol SO₂ and 2.00 mol O₂ if there was a 100% yield of SO₃, then 2 moles of SO₃ would be produced and 1.00 mol of O₂ would remain.
With a 71.0% yield, there are only 2*0.71 = 1.42 mol SO₃, the moles of SO₂ that didn't react would be 2 - 1.42 = 0.58; and the moles of O₂ that didn't react would be 2 - 1.42/2 = 1.29.

The total number of moles is 1.42 + 0.58 + 1.29 = 3.29. With that value we can calculate the molar fraction (X) of each component:

XSO₂ = 0.58/3.29 = 0.176

XO₂ = 1.29/3.29 = 0.392

XSO₃ = 1.42/3.29 = 0.432

The partial pressure of each gas is equal to the total pressure (PT) multiplied by the molar fraction of each component.

p(SO₂) = 0.176 * PT

p(O₂) = 0.392 * PT

p(SO₃) = 0.432 * PT

Rewriting KP and solving for PT:

$\frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}=0.13\\\frac{(0.432*P_{T})^{2}}{(0.176*P_{T})^{2}(0.392*P_{T})} =0.13\\\frac{0.1866*P_{T}^{2}}{0.0121*P_{T}^{3}} =0.13\\\frac{15.369}{P_{T}}=0.13\\P_{T}=118.22 atm$