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Deffense [45]
2 years ago
9

Calculate the theoretical yield for the bromination of both stilbenes

Chemistry
1 answer:
podryga [215]2 years ago
4 0

Answer:

cinnamic acid - 150 mg

cis-stilbene - 100 μL

trans- stilbene - 100 mg

pyridinium tribromide - 200-385 mg

For this data:

moles of cinnamic acid = 0.150 g/148.16 g/mol = 0.001 mols

Theoretical mass of dibromoproduct formed = 0.001 mol x 307.97 g/mol = 0.312 g

cis-stilbene (100 ul = 0.1 ml)

moles of cis-stilbene = 0.1 ml x 1.01 g/mol/180.25 g/mol = 0.00056 mols

Theoretical mass of dibromoproduct formed = 0.00056 mol x 340.05 g/mol = 0.19 g

trans-stilbene

moles of tran-stilbene = 0.1 g/180.25 g/mol = 0.00055 mols

Theoretical mass of dibromoproduct formed = 0.00055 mol x 340.05 g/mol = 0.19 g

Explanation:

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How many grams of NH3 can be produced from 2.30 mol of N2 and excess H2.
MrRa [10]
<h3>Answer:</h3>

78.34 g

<h3>Explanation:</h3>

From the question we are given;

Moles of Nitrogen gas as 2.3 moles

we are required to calculate the mass of NH₃ that may be reproduced.

<h3>Step 1: Writing the balanced equation for the reaction </h3>

The Balanced equation for the reaction is;

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<h3>Step 2: Calculating the number of moles of NH₃</h3>

From the equation 1 mole of nitrogen gas reacts to produce 2 moles of NH₃

Therefore, the mole ratio of N₂ to NH₃ is 1 : 2

Thus, Moles of NH₃ = Moles of N₂ × 2

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Mass = Moles × molar mass

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Mass = 4.6 moles × 17.031 g/mol

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