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DanielleElmas [232]
3 years ago
12

You are given a solution containing a pair of enantiomers (A and B). Careful measurements show that the solution contains 98% A

and 2% B. What is the ee of this solution
Chemistry
1 answer:
Andreyy893 years ago
8 0

Answer:

ee = 96%

Explanation:

Enantiomeric excess, ee, is a way to express a mixture that is not enantiomerically pure. It is defined as 100 times the ratio between the  differences of amounts of enantiomers and the total amunt. that is:

ee = |A-B|/ A+B * 100

ee = |98%-2%| / 98+2 * 100

<h3>ee = 96%</h3>
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2C2H6 + 8O2 = 4CO2 + 6H2O
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How many moles of water can be produced with 4.3 moles of H2 and 5.6 moles of O2? Which reactant is limiting? How many moles of
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Answer:

Hydrogen H₂ will be the limiting reagent.

The excess reactant that will be left after the reaction is 3.45 moles.

4.3 moles of water can be produced.

Explanation:

The balanced reation is:

2 H₂ + O₂ → 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

  • H₂: 2 moles
  • O₂: 1 mole
  • H₂O: 2 moles

To determine the limiting reagent, you can use a simple rule of three as follows: if by stoichiometry 1 mole of O₂ reacts with 2 moles of H₂, how much moles of H₂ will be needed if 5.6 moles of O₂ react?

moles of H_{2} =\frac{5.6 moles of O_{2} *2 mole of H_{2} }{1 mole of O_{2}}

moles of H₂= 11.2 moles

But 11.2 moles of H₂ are not available, 4.3 moles are available. Since you have less moles than you need to react with 5.6 moles of O₂, <u><em>hydrogen H₂ will be the limiting reagent</em></u> and oxygen O₂ will be the excess reagent.

Then you can apply the following rules of three:

  • If by reaction stoichiometry 2 moles of H₂ react with 1 mole of O₂, 4.3 moles of H₂ will react with how many moles of O₂?

moles of O_{2} =\frac{1 mole of O_{2} *4.3 mole of H_{2} }{2 mole of O_{2}}

moles of O₂= 2.15 moles

The excess reactant that will be left after the reaction can be calculated as:

5.6 moles - 2.15 moles= 3.45 moles

<u><em>The excess reactant that will be left after the reaction is 3.45 moles.</em></u>

  • If by reaction stoichiometry 2 moles of H₂ produce 2 moles of H₂O, 4.3 moles of H₂ produce how many moles of H₂O?

moles of H_{2}O =\frac{2 moles of H_{2}O *4.3 mole of H_{2} }{2 mole of H_{2}}

moles of H₂O= 4.3 moles

<u><em>4.3 moles of water can be produced.</em></u>

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