During a reaction, a halogen chemically combines with a Group 1 element to form a compound. how does this reaction occur, based
2 answers:
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Answer:
a. HCl.
b. 0.057 g.
c. 1.69 g.
d. 77 %.
Explanation:
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In this case, since the reaction between magnesium and hydrochloric acid is:
Whereas there is 1:2 mole ratio between them.
a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:
Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.
b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:
c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:
Thus, the mass of excess magnesium turns out:
d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:
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Answer:
See explanation and picture below
Explanation:
First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.
In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.
In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.
