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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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If the temperature of the sample of gas increases to the given value, the volume also increases to 600mL.

<h3>What is Charles's law?</h3>

Charles's law states that "the volume occupied by a definite quantity of gas is directly proportional to its absolute temperature.

It is expressed as;

V₁/T₁ = V₂/T₂

Given the data in the question;

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V₁/T₁ = V₂/T₂

V₂ = V₁T₂ / T₁

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V₂ = 60000mLK / 100K

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How many grams of NH3 can be produced from 2.51 mil of N2 and excess H2 ?
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Answer:

85.34g of NH3

Explanation:

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The balanced equation for the reaction. This is given below:

N2 + 3H2 —> 2NH3

Step 2:

Determination of the number of moles of NH3 produced by the reaction of 2.51 moles of N2. This is illustrated below:

From the balanced equation above,

1 mole of N2 reacted to produce 2 moles of NH3.

Therefore, 2.51 moles of N2 will react to produce = (2.51 x 2)/1 = 5.02 moles of NH3.

Therefore, 5.02 moles of NH3 is produced from the reaction.

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Conversion of 5.02 moles of NH3 to grams. This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17g/mol

Number of mole of NH3 = 5.02 moles

Mass of NH3 =..?

Mass = mole x molar Mass

Mass of NH3 = 5.02 x 17

Mass of NH3 = 85.34g

Therefore, 85.34g of NH3 is produced.

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