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Savatey [412]
3 years ago
7

If a mixture contains 75% of one compound and 25% of its enantiomer, what is the e.e. of the mixture

Chemistry
1 answer:
VMariaS [17]3 years ago
7 0

Answer: The enantiomeric excess ee is 50%

Explanation:

Given that;

compound A = 75%, B = 25%

ee = ?

so

ee = [( A - B) / ( A + B)] × 100

= [( 75 - 25) / ( 75 + 25)] × 100

= [50 / 100] × 100

= 0.5 × 100

= 50%

Therefore enantiomeric excess ee is 50%

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Read 2 more answers
2.0L of hydrogen gas is mixed with 3.0L of nitrogen gas at STP in a rigid 5.0L vessel. A reaction occurs, producing ammonia gas
IrinaK [193]

Answer:

Moles NH₃: 0.0593

0.104 moles of N₂ remain

Final pressure: 0.163atm

Explanation:

The reaction of nitrogen with hydrogen to produce ammonia is:

N₂ + 3 H₂ → 2 NH₃

Using PV = nRT, moles of N₂ and H₂ are:

N₂: 1atmₓ3.0L / 0.082atmL/molKₓ273K = 0.134 moles of N₂

H₂: 1atmₓ2.0L / 0.082atmL/molKₓ273K = 0.089 moles of H₂

The complete reaction of N₂ requires:

0.134 moles of N₂ × (3 moles H₂ / 1 mole N₂) = <em>0.402 moles H₂</em>

That means limiting reactant is H₂. And moles of NH₃ produced are:

0.089 moles of H₂ × (2 moles NH₃ / 3 mole H₂) = <em>0.0593 moles NH₃</em>

Moles of N₂ remain are:

0.134 moles of N₂ - (0.089 moles of H₂ × (1 moles N₂ / 3 mole H₂)) = <em>0.104 moles of N₂</em>

And final pressure is:

P = nRT / V

P = (0.104mol + 0.0593mol)×0.082atmL/molK×273K / 5.0L

<em>P = 0.163atm</em>

7 0
2 years ago
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