I have 3NaBH4 + 4BF3.Et2O to form 2B2H6 th number of moles of NaBH4 is 0.0368mol and the number of moles of BF3.Et2O is 0.0162mol accoring to the ratio of reaction 3:4 the limiting reactant is BF3.Et2O so its the one used to find number of moles of dibroane (B2H6)? and then in the actual reaction its 3cyclohexene (0.0593mo) + b2h6 + NaOH(0.348) + H2O2(0.256mol) to form 3cyclohexanol since B2H6 is limiting its the one used to find number of moles of cyclohexanol?
I don't see a question here.
I mean do you its right?
The first part is right; i.e., the BF3.Et2O is the limiting reagent (LR). I don't know about the last part of that with the cyclohexene. What you must do for that part is to determine how much of the B2H6 is available for the cyclohexene reaction. But don't forget you see mols NaOH there too so you must determine, with the new amount of B2H6 AND the amount of NaOH and cyclohexene which is the LR.
how can cyclohexene be limiting reagent its number of moles = 0.0593mol and b2h6 as calculated earlier 0.0162mol and they are in a ratio of 1:3 so b2h6 is limiting reactant therefore b2h6 is the one that contributes for the formation of cyclohexanol? NaOH and H2O2 are not limiting in comparsion to b2h6 and cyclohexene as well
First I didn't say cyclohexene was the LR. You may have calculated that and if that's the result you get you go with it. It is important for you to realize that if you are calculating this in two consecutive steps that the second step depends upon how much B2H6 is formed and how much of the other reagents are present and determine the LR for that step also.
So you said B2H6 is the one reacting with cyclohexene is it not BH3 if its B2H6 reacting than its right but I was taught that BH3 is the one that reacts so the splitting of diborane (B2H6) to two molecules of borane has to be considered and the moles ratio to be considered as well B2H6➡️2BH3?
When cyclohexene reacts with BH3 it forms trialklyborane boron in the middle with three cyclohexane ring around it and then it react with NaOH and Hydrogen leroxide (H2O2) to form cyclohexanol so I have consider sodium hydroxide or hydrogen per oxide aa th LR here not BH3 is that right?
Yes, you are correct in determining the limiting reactant in the first reaction. According to the given ratio of the reaction, the moles of NaBH4 (0.0368 mol) and BF3.Et2O (0.0162 mol) are in a 3:4 ratio. This means that for every 3 moles of NaBH4, you need 4 moles of BF3.Et2O to react completely.
Since you have fewer moles of BF3.Et2O than NaBH4, BF3.Et2O is the limiting reactant. It will be completely consumed in the reaction, and you will not have any excess left over after the reaction is complete.
To find the number of moles of diborane (B2H6), you need to use the stoichiometry of the reaction. From the balanced equation, you can see that 4 moles of BF3.Et2O react to produce 2 moles of B2H6.
First, calculate the moles of B2H6 using the moles of BF3.Et2O:
Number of moles of B2H6 = (0.0162 mol BF3.Et2O) x (2 mol B2H6 / 4 mol BF3.Et2O)
Number of moles of B2H6 = 0.0081 mol
Therefore, the number of moles of B2H6 produced in the first reaction is 0.0081 mol.
In the second reaction, since B2H6 is the limiting reactant, it's the one that determines the number of moles of cyclohexanol.
Looking at the balanced equation, you can see that 1 mole of B2H6 reacts to produce 3 moles of cyclohexanol.
To find the number of moles of cyclohexanol, use the moles of B2H6 obtained from the first reaction:
Number of moles of cyclohexanol = (0.0081 mol B2H6) x (3 mol cyclohexanol / 1 mol B2H6)
Number of moles of cyclohexanol = 0.0243 mol
Therefore, the number of moles of cyclohexanol produced in the second reaction is 0.0243 mol.