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Sladkaya [172]
3 years ago
14

You plan to separate a polar and a non-polar compound using normal phase column chromatography technique. Methylene chloride (CH

2Cl2) and hexanes (C6H14)are the solvents to be used to elute the column. Which solvent should be used first
Chemistry
1 answer:
marissa [1.9K]3 years ago
3 0

Answer:

Hexane should be used first.

Explanation:

Chromatography is a method of separating the constituents of a mixture by taking advantage of their different rates of movement in a solvent over an adsorbent medium. Chromatography is a means of separation and analysis that utilises fractional separation. It is based on the principle  that if a fluid containing a number of substances is allowed to pass though an adsorbent medium, the different substances in the fluid may travel at different rates and be separated.

The rate of movement depends on the relative affinities of the constituents for the solvent and adsorbent medium. i.e solutes which are weakly adsorbed by the adsorbent medium are easily redissolved by the ascending solvent and quickly travel up the adsorbent medium. In addition to that , solutes which are very soluble in the solvent move up at a faster rate than those which  are not soluble.

In column Chromatography;

A non-polar solvent should be initiated and applied first. This is because , in a column chromatography, a non-polar compound will be removed at first then later polar compound.

Assuming a polar compound is used first, the polar compounds will be removed alongside with all the non-polar compounds.

From the two Compounds given;

We know that :

Hexane is a non-polar compound and Methylene chloride  is a polar compound. As such, Hexane should be used first.

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Answer:

a) 1.39 g ; b) O₂ is limiting reactant,  NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.  

MM:        17.03    32.00     30.01

              4NH₃  +  5O₂ ⟶ 4NO + 6H₂O

Mass/g:    1.5        1.85

2. Calculate the moles of each reactant  

\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

\text{Moles of NO} =  \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}

7. Calculate the mass of NH₃ reacted

\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

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