Answer:
12 moles of H₂O are formed in this combustion.
Explanation:
First of all, think the reaction:
2CH₃OH (l) + 3O₂ (g) → 2CO₂ (g) + 4H₂O (g)
Ratio in the reactants is 2:3, so 2 mol of methanol need 3 mol of oxygen to react. Then 8 mol of CH₃OH, will need (8.3)/2 = 12 moles of O₂
We have 9 moles of O₂, so this is the limiting reactant.
3 mol of oxygen produce 4 mol of water
Then, 9 mol of oxygen will produce ( 9 .4)/3 = 12 moles
Answer:
Reagents: 1) 
2) 
, 
Mechanism: Hydroboration
Explanation:
In this case, we have a <u>hydration of alkene</u>s reaction. But, in this example, we have an <u>anti-Markovnikov reaction</u>. In other words, the "OH" is added in the least substituted carbon. Therefore we have to choose an anti-Markovnikov reaction: <u>"hydroboration"</u>.
The <u>first step</u> of this reaction is the addition of borane () to the double bond. Then in the <u>second step</u>, we have the deprotonation of the hydrogen peroxide, to obtain the peroxide anion. In the <u>third step</u>, the peroxide anion attacks the molecule produced in the first step to produce a complex compound in which we have a bond "
". In <u>step number 4</u> we have the migration of the C-B bond to oxygen. Then in <u>step number 5</u>, we have the attack of on the 
to produce an alkoxide. Finally, the water molecule produce in step 2 will <u>protonate</u> the molecule to produce the alcohol.
See figure 1
I hope it helps!
1.) Gas given off
2.) precipitate formed
3.) large amount of heat given off
4.) Color change
