how many grams of H2 are needed to produce 14.34 g of NH3?

please help, this hw is due in a few hours

Answer:

pH = 8.75

Explanation:

100.0mL of a 0.8108M of a butanoic acid (HC₃H₇CO₂, pKa 4.82) solution is titrated with 0.0520M KOH.

The reaction is:

HC₃H₇CO₂ + KOH → H₂O + KC₃H₇CO

Moles of butanoic acid are:

0.1000L × (0.8108mol / L) = 0.08108 moles of butanoic

For a complete reaction, volume of KOH must be added is the volume in which 0.08108 moles of KOH are added, that is:

0.08108 mol × (L / 0.0520mol) = 1.56L of KOH.

Total volume in equilibrium is 1.56L + 0.10L = 1.66L

That means concentration of butanoic acid is:

0.08108 mol / 1.66L = 0.04884M HC₃H₇CO₂

At equivalence point, there is just C₃H₇CO⁻ in solution

Kb of butanoic acid is:

C₃H₇CO⁻ + H₂O ⇄ HC₃H₇CO₂ + OH⁻

Kb = Kw / Ka

Ka = 10^-pKa

Ka = 1.51x10⁻⁵

Kb = 1x10⁻¹⁴ / 1.51x10⁻⁵ = 6.61x10⁻¹⁰

The equilibrium of Kb is:

Kb = 6.61x10⁻¹⁰ =  [HC₃H₇CO₂] [OH⁻] / [C₃H₇CO⁻]

As at equivalence point there is just C₃H₇CO⁻, the equilibrium concentrations are:

[C₃H₇CO⁻] = 0.04884M - X

[HC₃H₇CO₂] = X

[OH⁻] = X

Replacing in Kb:

6.61x10⁻¹⁰ =  X² / [0.04884M - X]

0 =  X² + 6.61x10⁻¹⁰X - 3.23x10⁻¹¹

Solving for X:

X = -5.68x10⁻⁶ → False solution. There is no negative concentrations

X = 5.683x10⁻⁶ → Right solution.

As [OH⁻] = X, [OH⁻] = 5.683x10⁻⁶.

pOH = - log [OH⁻]

pOH = 5.245

pH = 14 - pOH

Answer:c

Explanation: rusting, burning and fuzzing are all examples of chemical reactions/changes.