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Alja [10]
3 years ago
15

Write the balanced NET IONIC equation for the reaction that occurs when ammonium nitrate and potassium hydroxide are combined. N

H4 OH- This reaction is classified as . A. Strong Acid Strong Base B. Weak Acid Strong Base C. Strong Acid Weak Base D. Weak Acid Weak Base The extent of this reaction is: . A. ... Below 50% B. ... 50% C. ... Above 50% D. ... 100%
Chemistry
1 answer:
vova2212 [387]3 years ago
8 0

Answer:

Net Ionic equation

NH₄⁺ + OH⁻ → NH₃ + H₂O

Option B is correct.

Weak Acid Strong Base

Check Explanation for the extent of the reaction.

Explanation:

Ammonium nitrate = NH₄NO₃

Potassium Hydroxide = KOH

Ammonium salts combine with alkalis to liberate NH₃ and form water.

The two reactants combine to give

NH₄NO₃ + KOH → KNO₃ + NH₃ + H₂O

In ionic form,

- NH₄NO₃ exists as NH₄⁺ and NO₃⁻

- KOH exists as K⁺ and OH⁻

- KNO₃ as K⁺ and NO₃⁻

And NH₃ and H₂O stay as they are, as per covalent compounds.

So, we have

NH₄⁺ + NO₃⁻ + K⁺ + OH⁻ → K⁺ + NO₃⁻ + NH₃ + H₂O

Eliminating the ions that exist on both sides, we have the net ionic equation to be

NH₄⁺ + OH⁻ → NH₃ + H₂O

which shows that this reaction is essentially a neutralization reaction in which the Bronsted Lowry acid, NH₄⁺, loses its proton to the base, OH⁻ and gives conjugate base, NH₃ and conjugate acid, H₂O.

This reaction is classified as a Weak acid versus Strong Base reaction as NH₄⁺ is from a Weak acid and OH⁻ is from a strong base.

Since this reaction is between a Weak base and a strong acid, the ionization isn't expected to be 100%, Hence, the extent of this reaction will be any option that is not 100%, a couple pieces of information might be required for the correct estimate, but above 50% seems correct.

Hope this Helps!!!

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1) Endothermic

2) Heat absorbed by reaction = 2511.6J

3)ΔH_rxn = 18.2 kJ/ mol

Explanation:

Mass of KCl = 10.3 g

Moles of KCl = mass of KCl / molar mass of KCl

= 10.3 g / 74.5513 g/mol

=0.138 mol

Mass of water, m = 200 g

Specific heat of water, C= 4.186 Jg^-1°C^-1

T_initial = 22.0C

T_final = 19.0°C

ΔT = T_final - T_initial

=19.0°C - 22.0°C

= -3.0°C

Heat absorbed = Heat released

q_rxn =-q_water

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q_water = m CΔT

substitute

q_water = 200g * 4.186J g^-1 °C^-1 *(-3.0°C)

= -2511.6J

Heat absorbed by reaction = 2511.6J

q_rxn = -q_water = 2511.6J

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