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Identify the products represented by letters A-F to complete each chemical equation, then write the net ionic equations for the reactions including the physical states of reactants and products. Use the standard enthalpies of formation to calculate the changes in the standard enthalpies of reaction.
a) NaOH(aq)+HCl(aq)⟶A(aq)+B(l)
A=

B=

Balanced net ionic equation:
Δ0rxn=


b) NaOH(aq)+NH4Cl(aq)⟶C(aq)+D(aq)+E(l)
C=
D=
E=
Balanced net ionic equation:
Δ0rxn=


c) HCl(aq)+NH3(aq)⟶F(l)
F=

Balanced net ionic equation:


Δ0rxn=

Respuesta :

The product of a chemical reaction is what is remaining at the end of the reaction. The combination between reactant to form products can happen in a variety of ways including combination of ions in a solution to form new compounds

The correct compounds representing the variables and the net ionic equations are as follows:

a) A = NaCl

B = H₂O

Balanced net ionic equation: OH⁻(aq) + H⁺(aq) → H₂O (l)

ΔHrxn = Negative

b) C = NH₃

D = NaCl(aq)

E = H₂O(l)

Balanced net ionic equation: OH⁻(aq) + NH₄⁺(aq) → NH₃(aq) + H₂O(l)

ΔHrxn = Negative

c) F = NH₄Cl(aq)

Balanced net ionic equation: H⁺(aq) + NH₃(aq) → NH₄⁺(aq)

ΔHrxn = Negative

The reason the above values are correct are as follows:

a) Part (i)

The given chemical equation is

NaOH(aq) + HCl(aq) → A(aq) + B(l)

The chemical equation is the reaction between sodium hydroxide NaOH (aq) and hydrochloric acid HCl (aq), the product of the reaction is sodium chloride and water as follows;

NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l) + Heat

Therefore, we have;

A = NaCl, and B = H₂O

Part (ii)

The ionic equation is given as follows;

Na⁺(aq) + OH⁻(aq) + H⁺(aq) + Cl⁻(aq) → Na⁺(aq) + Cl⁻(aq) + H₂O (l)

Given that Na⁺(aq) and Cl⁻(aq) are spectator ions in the reaction, we have the balanced net ionic equation given as follows;

Balanced net ionic equation: OH⁻(aq) + H⁺(aq) → H₂O (l)

Part (iii)

ΔHrxn = Negative

b) Part (i)

The given chemical equation is NaOH(aq) + NH₄Cl(aq) → C(aq) + D(aq) + E(l)

The given reaction is between sodium hydroxide, NaOH(aq) and ammonium chloride, NH₄Cl(aq) and the products of the reaction are dissolved ammonia NH₃(aq), sodium chloride NaCl(aq), and water H₂O(l)

Therefore, we have;

NaOH(aq) + NH₄Cl(aq) → NH₃(aq) + NaCl (aq) + H₂O(l) + Heat

Therefore;

C = NH₃, D = NaCl(aq), E = H₂O(l)

Part (ii)

The ionic equation for the reaction is given as follows;

Na⁺(aq) + OH⁻(aq) + NH₄⁺(aq) + Cl⁻(aq) → NH₃(aq) + Na⁺(aq) + Cl⁻(aq) + H₂O(l)

The Na⁺(aq) and Cl⁻(aq) occur in both the reactants and products, therefore, they cancel out as spectator ions and we have;

Balanced net ionic equation: OH⁻(aq) + NH₄⁺(aq) → NH₃(aq) + H₂O(l)

Part (iii)

ΔHrxn = Negative

c) Part (i)

HCl(aq) + NH₃(aq) → F(l)

The given reaction is between hydrochloric acid and aqueous ammonia which has the product as ammonium chloride, NH₄Cl

Therefore, the reaction is given as follows;

HCl(aq) + NH₃(aq) → NH₄Cl(aq) + Heat

∴ F = NH₄Cl(aq)

Part (ii)

The ionic equation is given as follows;

H⁺(aq) + Cl⁻(aq) + NH₃(aq) → NH₄⁺(aq) + Cl⁻(aq)

Therefore, the chlorine ion Cl⁻(aq) occurs in both the reactant and product and can therefore, can be canceled out, and we have;

Balanced net ionic equation: H⁺(aq) + NH₃(aq) → NH₄⁺(aq)

Part (iii)

ΔHrxn = Negative

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