Addition of water to an alkyne gives a keto‑enol tautomer product and that is the product changed into 2-pentanone, then the alkyne need to had been 1-pentyne. 2-pentyne might have given a combination of 2- and 3-pentanone.
<h3>
What is the keto-enol means in tautomer?</h3>
They carries a carbonyl bond even as enol implies the presence of a double bond and a hydroxyl group. The keto-enol tautomerization equilibrium is depending on stabilization elements of each the keto tautomer and the enol tautomer.
- The enol that could provide 2-pentanone might had been pent-1- en - 2 -ol. Because an equilibrium favors the ketone so greatly, equilibrium isn't an excellent description.
- If the ketone have been handled with bromine, little response might be visible because the enol content material might be too low.
- If a catalyst have been delivered, NaOH for example, then formation of the enolate of pent-1-en - 2 - ol might shape and react with bromine.
- This might finally provide a bromoform product. Under acidic conditions, the enol might desire formation of the greater substituted enol constant with alkene stability.
Answer:
3.93 mol
Explanation:
The balanced equation is given as;
2 Fe₂S₃+ 9 O₂ --> 2 Fe₂O₃ + 6 SO₂
From the reaction;
2 mol of Fe₂S₃ reacts with 9 mol of O₂ to form 6 mol of 6 SO₂
This means;
1 mol of Fe₂S₃ requires 9/2 mol of O₂
Also,
1 mol of O₂ requires 1/9 mol of Fe₂S₃
In the question, we have;
1.31 moles of Fe2S3 and 22.8 moles of O2
The limiting reactant which determine how much of the product formed is; Fe₂S₃ because O₂ is in excess.
The relationship between Fe₂S₃ and SO₂ is;
2 mol Fe₂S₃ produces 6 mol of SO₂
1.31 mol of Fe₂S₃ would produce x mol ?
2 = 6
1.31 = x
x = 6 * 1.31 / 2 = 3.93 mol
During a chemical reaction nb of atoms and mass is always conserved , so mass of Reactants is equal mass of Products.
The reaction H2SO4 + 2 NaOH -> Na2SO4 + 2H2O applies to an acid-base titration.
moles NaOH = c · V = 0.2423 mmol/mL · 32.23 mL = 7.809329 mmol
moles H2SO4 = 7.809329 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 3.9046645 mmol
Hence
[H2SO4]= n/V = 3.9046645 mmol / 37.21 mL = 0.1049 M
The answer to this question is [H2SO4] = 0.1049 M
