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The question is incomplete, here is the complete question:
While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures.
A chemical engineer studying this reaction fills a 1.5 L flask at 12°C with 1.8 atm of ethylene gas and 4.7 atm of water vapor. When the mixture has come to equilibrium she determines that it contains 1.16 atm of ethylene gas and 4.06 atm of water vapor.
The engineer then adds another 1.2 atm of ethylene, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.
<u>Answer:</u> The partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm
<u>Explanation:</u>
We are given:
Initial partial pressure of ethylene gas = 1.8 atm
Initial partial pressure of water vapor = 4.7 atm
Equilibrium partial pressure of ethylene gas = 1.16 atm
Equilibrium partial pressure of water vapor = 4.06 atm
The chemical equation for the reaction of ethylene gas and water vapor follows:
<u>Initial:</u> 1.8 4.7
<u>At eqllm:</u> 1.8-x 4.7-x
Evaluating the value of 'x'
The expression of for above equation follows:
Putting values in above expression, we get:
When more ethylene is added, the equilibrium gets re-established.
Partial pressure of ethylene added = 1.2 atm
<u>Initial:</u> 2.36 4.06 0.64
<u>At eqllm:</u> 2.36-x 4.06-x 0.64+x
Putting value in the equilibrium constant expression, we get:
Neglecting the value of x = 13.41 because equilibrium partial pressure of ethylene and water vapor will become negative, which is not possible.
So, equilibrium partial pressure of ethanol = (0.64 + x) = (0.64 + 0.363) = 1.003 atm
Hence, the partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm
Answer:
a) 1.39 g ; b) O₂ is limiting reactant, NH₃ is excess reactant; c) 0.7 g
Explanation:
We have the masses of two reactants, so this is a limiting reactant problem.
We will need a balanced equation with masses, moles, and molar masses of the compounds involved.
1. Gather all the information in one place with molar masses above the formulas and masses below them.
MM: 17.03 32.00 30.01
4NH₃ + 5O₂ ⟶ 4NO + 6H₂O
Mass/g: 1.5 1.85
2. Calculate the moles of each reactant
3. Calculate the moles of NO we can obtain from each reactant
From NH₃:
The molar ratio is 4 mol NO:4 mol NH₃
From O₂:
The molar ratio is 4 mol NO:5 mol O₂
4. Identify the limiting and excess reactants
The limiting reactant is O₂ because it gives the smaller amount of NO.
The excess reactant is NH₃.
5. Calculate the mass of NO formed
6. Calculate the moles of NH₃ reacted
The molar ratio is 4 mol NH₃:5 mol O₂
7. Calculate the mass of NH₃ reacted
8. Calculate the mass of NH₃ remaining
Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃