For a given sample of ch3oh, the enthalpy change during the reaction is 82.3 kj . how many grams of methane gas are produced?

1 answer:

8 0

The equation is as follows;
2CH3OH(g) = 2CH4(g) + O2(g), ΔH= +252.8 kJ
From the equation; for the reaction to produce 2 moles of methane (32g) an energy of 252.8 kJ is released.
Therefore; for an energy of 82.3 kJ the number of moles that will be produced will be; = (2×82.3)/252.3
= 0.6524 moles
which is equivalent to 0.6524 × 16 = 10.438 g
There, the mass of CH4 produced will be 10.438 g

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Give regents and mechanism

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Answer:

Reagents: 1) $BH_3$ 2) $H_2_O2$ , $OH^-$

Mechanism: Hydroboration

Explanation:

In this case, we have a hydration of alkenes reaction. But, in this example, we have an anti-Markovnikov reaction. In other words, the "OH" is added in the least substituted carbon. Therefore we have to choose an anti-Markovnikov reaction: "hydroboration".

The first step of this reaction is the addition of borane ( $BH_3$ ) to the double bond. Then in the second step, we have the deprotonation of the hydrogen peroxide, to obtain the peroxide anion. In the third step, the peroxide anion attacks the molecule produced in the first step to produce a complex compound in which we have a bond " $BH_2-O-OH$ ". In step number 4 we have the migration of the C-B bond to oxygen. Then in step number 5, we have the attack of $OH^-$ on the $O-BH_2$ to produce an alkoxide. Finally, the water molecule produce in step 2 will protonate the molecule to produce the alcohol.