A CHEMILUMINESCENT COMPOUND: LUMINOL Discussion describes the production of ligh

Question

A CHEMILUMINESCENT COMPOUND: LUMINOL Discussion describes the production of light as a result of a chemical reaction. oerally produce a product in an electronically excited state. As the Chemiluminescence describes These reactions generally different s to its ground state, a photon is emitted, and light is produced. Many known ors have adopted these reactions for specific purposes. The most well compoundnism would probably be the American firefly. The firefly makes use of a understooderin, but the complete reaction to produce light is not completely organisms organism In this exp-nohth diamide from 3-nitrophthalic diamides experiment, luminol (5-aminophthalhydrazide) is prepared by formation of a cyclic 3-nitrophthalic acid and hydrazine, followed by reduction of the nitro group. NH,NII 3-nitrophthalic acid Chemical Formula CaHyNO 5-nitro-2,3-dihydrophthalazine-1,4-dione s "luminol" Molecular Weight: 211.13 g/mol Chemical Formula: CHN O Molecular Weight: 177.16 g/mol n alkaline solution, luminol is converted to its dianion, and can be oxidized to an intermediate that is hemiluminescent. The light produced is a blue-green color, and non-thermal in origin (i.e. no heat is oduced) 0 NH +20H +02 NH intersystem crossing from triplet to singlet excited state NH2 O NH2

Explanation / Answer

To determine a limiting reagent ,first we have to write a balanced equation.

C8H5NO6 + NH2NH2 ------> C8H5N3O4 +2 H2O

C8H5N3O4 + Na2S2O4  -------->C8H7N3O2

stoiciometry of the reaction is 1:1

Now, we will calculate no. Of moles of reactant used:

1) C8H5NO6 used is 1 g;

No. Of moles = 1 g ×( 1 mole/ 211.13 g) = 0.0047 mole

2) NH2NH2 used is 2 mL

No. Of moles = volume of hydrazine × density of hydrazine× (1mole/ molecular weight of hydrazine)

= 2 mL×( 1.0045 g/mL) ×(1 mole/32.05 g) = 0.062 moles

3) Na2S2O4 used is 3 g

No. Of moles Na2S2O4 = 3 g x (1 mole / 174.11g) = 0.017 moles

The mole ratio of rxn 1 is 1:1 thus limiting reagent is C8H5NO6,, sine hydrazine is present in excess

Also in rxn 2 molar ratio between reactant is 1:1 , so C8H5N3O4 i a limiting reagent

Now we will calculate yield of reaction

Theoritical yield:

0.0047 mole of C8H5NO6 × ( 1 mole C8H5N3O4 / 1 mole C8H5NO6) = 0.0047 mole of C8H5N3O4

similarly, 0.0047 moles of C8H5N3O4 gives 0.0047 moles of luminol

0.0047 moles of luminol = 0.0047 mole luminol × (177.16 g luminol/ mole) = 0.8326 g luminol.....this is the theoritical yield.

To calculate percent yield = ( experimental yield / theoritical yield) ×100

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