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nevsk [136]
3 years ago
11

Calcium cyclamate, Ca(C₆H₁₁NHSO₃)₂, is an artificial sweetener used in many countries around the world but is banned in the Unit

ed States. It can be purified industrially by converting it to the barium salt through reaction of the acid HC₆H₁₁NHSO₃ with barium carbonate (BaCO₃), treatment with aqueous sulfuric acid (H₂SO₄)—barium sulfate is very insoluble—followed by neutralization with calcium hydroxide solution (Ca(OH)₂). Write the balanced equations for these reactions.
Chemistry
1 answer:
xeze [42]3 years ago
3 0

Answer:

2HC₆H₁₁NHSO₃ + BaCO₃ → Ba(C₆H₁₁NHSO₃)₂ + H₂CO₃;

Ba(C₆H₁₁NHSO₃)₂ + H₂SO₄ → BaSO₄ + 2HC₆H₁₁NHSO₃;

2HC₆H₁₁NHSO₃ + Ca(OH)₂ → Ca(C₆H₁₁NHSO₃)₂ + 2H₂O

Explanation:

First, let's see the reactants for the first reaction and how they dissociate:

HC₆H₁₁NHSO₃ → H⁺ + C₆H₁₁NHSO₃⁻

BaCO₃ → Ba²⁺ + CO₃²⁻ (Barium is from group 2, so its cation has charge +2)

So, to form the products, the cation of one will join the anion of others. The amount of the cation will be the charge of the anion, and the amount of the anion will be the charge of the cation:

H⁺ + CO₃²⁻ → H₂CO₃

Ba²⁺ + C₆H₁₁NHSO₃⁻ → Ba(C₆H₁₁NHSO₃)₂

The reaction then is:

HC₆H₁₁NHSO₃ + BaCO₃ → Ba(C₆H₁₁NHSO₃)₂ + H₂CO₃

The number of elements must be the same on both sides, so the balanced equation is

2HC₆H₁₁NHSO₃ + BaCO₃ → Ba(C₆H₁₁NHSO₃)₂ + H₂CO₃

The treatment with H₂SO₄ will produce:

H₂SO₄ → 2H⁺ + SO₄⁻²

Ba(C₆H₁₁NHSO₃)₂ → Ba²⁺ + C₆H₁₁NHSO₃⁻

The balanced reaction will be then:

Ba(C₆H₁₁NHSO₃)₂ + H₂SO₄ → BaSO₄ + 2HC₆H₁₁NHSO₃

In the last step, HC₆H₁₁NHSO₃ will react with Ca(OH)₂

HC₆H₁₁NHSO₃ → H⁺ + C₆H₁₁NHSO₃⁻

Ca(OH)₂ → Ca²⁺ + 2OH⁻

The balance reaction will be:

2HC₆H₁₁NHSO₃ + Ca(OH)₂ → Ca(C₆H₁₁NHSO₃)₂ + 2H₂O

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andre [41]

<u>Answer:</u> The enthalpy change of the reaction is -243 J/mol

<u>Explanation:</u>

The heat released by the reaction is absorbed by the calorimeter and the solution.

The chemical equation used to calculate the heat released follows:

q=c\times \Delta T

where,

c = heat capacity of calorimeter = 1620 J/K

\Delta T = change in temperature = 0.150^oC=0.150K   (Change remains same)

Putting values in above equation, we get:

q=1620J/K\times 0.15K=243J

<u>Sign convention of heat:</u>

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

For the given chemical reaction:

S\text{ (monoclinic)}\rightarrow S\text{ (orthorhombic)}

We are given:

Moles of monoclinic sulfur = 1 mole

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\Delta H_{rxn}=\frac{q}{n}

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q = amount of heat released = -243 J

n = number of moles = 1 mole

\Delta H_{rxn} = enthalpy change of the reaction

Putting values in above equation, we get:

\Delta H_{rxn}=\frac{-243J}{1mol}=-243J/mol

Hence, the enthalpy change of the reaction is -243 J/mol

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