How do I figure out the limiting reagent for this experiment? Preparation of diphenylacetylene! Step 1:: trans-stilbene + pyridinium hydrobromide perbromide in glacial acetic acid -----> stilbene dibromide
Starting weight of trans-stilbene: 2.014g
Starting weight of pyridinium hydrobromide perbromide: 5.0g
Starting volume of glacial acetic acid: 40.0mL
Weight of the recovered product (stilbene dribromide): 3.709g
Help???
MW of trans-stilbene= 180.25g/mol
MW of pyridinium hydrobromide perbromide = 319.84g/mol
MW of stilbene dibromide= 340.05g/mol
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To determine the limiting reagent in the preparation of diphenylacetylene, you need to compare the moles of each reactant and their stoichiometric ratios. Here's how you can calculate it step-by-step:
Step 1: Calculate the moles of each reactant:
Moles of trans-stilbene = (mass of trans-stilbene) / (molecular weight of trans-stilbene)
= 2.014g / 180.25g/mol
= 0.0112 mol
Moles of pyridinium hydrobromide perbromide = (mass of pyridinium hydrobromide perbromide) /
(molecular weight of pyridinium hydrobromide perbromide)
= 5.0g / 319.84g/mol
= 0.0156 mol
Step 2: Determine the stoichiometric ratio between reactants:
From the balanced chemical equation, you can see that the ratio between trans-stilbene and pyridinium hydrobromide perbromide is 1:1.
Step 3: Compare the moles of the reactants:
Since the stoichiometric ratio is 1:1, whichever reactant has fewer moles will be the limiting reagent.
Moles of trans-stilbene = 0.0112 mol
Moles of pyridinium hydrobromide perbromide = 0.0156 mol
Since trans-stilbene has fewer moles, it is the limiting reagent.
Therefore, trans-stilbene is the limiting reagent in this experiment.
To determine the limiting reagent in a chemical reaction, you need to compare the amount of each reactant to the stoichiometry of the balanced equation. The limiting reagent is the reactant that will be completely consumed first, limiting the amount of product that can be formed.
Step 1: Write the balanced equation for the reaction:
trans-stilbene + pyridinium hydrobromide perbromide + glacial acetic acid → stilbene dibromide + other products
Step 2: Calculate the number of moles of each reactant.
Number of moles of trans-stilbene = Starting weight of trans-stilbene / Molecular weight of trans-stilbene
Number of moles of pyridinium hydrobromide perbromide = Starting weight of pyridinium hydrobromide perbromide / Molecular weight of pyridinium hydrobromide perbromide
Step 3: Determine the stoichiometry of the reaction.
From the balanced equation, we can see that the stoichiometry is 1:1:1, meaning 1 mole of trans-stilbene reacts with 1 mole of pyridinium hydrobromide perbromide to produce 1 mole of stilbene dibromide.
Step 4: Compare the moles of reactants to the stoichiometry.
Divide the number of moles of each reactant by the stoichiometric coefficient:
Moles of trans-stilbene / 1 = Moles of trans-stilbene
Moles of pyridinium hydrobromide perbromide / 1 = Moles of pyridinium hydrobromide perbromide
Step 5: Identify the limiting reagent.
The limiting reagent is the one with fewer moles compared to the stoichiometry. Whichever reactant has fewer moles is the limiting reagent.
Step 6: Calculate the theoretical yield of the product.
Using the balanced equation, the number of moles of the limiting reagent determines the number of moles of the product that can be formed.
Theoretical yield (in moles) = Moles of limiting reagent * Stoichiometric coefficient of the product
Finally, you can convert the theoretical yield into grams by multiplying it by the molecular weight of stilbene dibromide.
The limiting reagent is responsible for determining the maximum amount of product that can be produced. The reactant that is not the limiting reagent is in excess and will have some amount left over after the reaction completes.