What halide would undergo dehydrohalogenation to give 2-pentene as a pure product?
2 answers:
Answer:
The correct answer is 3-Chloropentane.
Explanation:
Dehydrohalogenation refers to a chemical reaction, which includes elimination of a hydrogen halide from a substrate. The reaction is generally related to the production of alkenes, however, it has various uses. Conventionally, the alkyl halides are the substrates for dehydrohalogenations. The alkyl halide must possess the tendency to produce an alkene, therefore, benzyl and methyl halides are not the ideal substrates.
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Explanation:
(a) It is given that magnesium is reacted with hydrochloric acid and the hydrogen evolved is collected at top. This means that hydrochloric acid will be present in a solution (HCl + Water) and the solvent will be water.
Due to evaporation some amount of water will have evaporated and would be present in vapor phase. Therefore, when the reaction occurs only hydrogen will not be present in vapor phase but, will be accompanied by water vapors as well .
Hence, Dalton's law the total pressure of the system will be sum of pressure exerted by hydrogen gas and pressure exerted by water vapors .
Let us assume that the partial pressure of hydrogen gas be ""
And, the partial pressure of water will be nothing but the vapor pressure of water,
Vapor pressure of water =
= 19.8 mm Hg
Total pressure of the system = 746 mm Hg
Total pressure =
746 = + 19.8
or, = 746-19.8
= 726.2 mm Hg
Hence, partial pressure of hydrogen gas is 726.2 mm Hg.
(b) To calculate volume at STP, we will first calculate at and 726.2 mm Hg and than convert it to STP conditions.
Therefore, to calculate volume at and 726.2 mm Hg we will make use of ideal gas law as follows.
P = 726.2 mm Hg
=
= 0.955 atm
T =
= 22+273.15 = 295.15 K
V = 31 ml = Litre
According to the ideal gas law ,
PV = nRT
where, P = pressure of the system ,
V = volume of the gas
N = number of moles
R = 0.0821 liter atm/mole K
T = Temperature
Hence, putting the given values into the above formula as follows.
N = moles
Now, the moles of hydrogen won't change. Therefore, let us calculate volume at STP of moles of hydrogen.
Now, at STP T = 273.15 K , P = 1 atm and N = moles
V = 0.027398 Litre
= (as 1 L = 1000 ml)
= 27.398 ml
Therefore, volume of hydrogen at STP is 27.398 ml .
(c) Now, we can write the the reaction for this case as follows.
As, weight of magnesium = 0.028 grams
Molar mass of magnesium = 24.3 grams/mole
Number of moles of magnesium =
=
= moles
Since, it can be seen from the reaction that
1 mole of Magnesium = 1 mole of hydrogen
and, moles of hydrogen = moles
= 0.001523 moles
Hence, theoretical number of moles of hydrogen that can be produced from 0.028 grams of Mg is 0.001523 moles
Answer:
191.11 grams of oxygen gas should be produced.
Explanation:
The balanced reaction is:
2 Al₂O₃ → 4 Al + 3 O₂
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Al₂O₃: 2 moles
- Al: 4 moles
- O₂: 3 moles
Being the molar mass of each compound:
- Al₂O₃: 102 g/mole
- Al: 27 g/mole
- O₂: 32 g/mole
By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Al₂O₃: 2 moles* 102 g/mole= 204 grams
- Al: 4 moles* 27 g/mole= 108 grams
- O₂: 3 moles* 32 g/mole= 96 grams
Then you can apply the following rule of three: if by stoichiometry 108 grams of aluminum are produced along with 96 grams of oxygen, 215 grams of aluminum are produced along with how much mass of oxygen?
mass of oxygen= 191.11 grams
<u><em>191.11 grams of oxygen gas should be produced.</em></u>