Question

Can someone pliz use the lab experiment n answer the question below. Especially calculating limiting reagent n theoretical yield. Pliz show well organize work

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 。

Answer 1

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 C₈H₅NO_6,, 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