Finding the moles required

[omzar]

ive been having a hard time with this question and any help would be appreciated.

If the percent yield in the reaction of methyl iodide (CH3I) with water to form methanol (CH3OH) and HI is 50%, how many moles of methyl iodide will be required to prepare 0.1 mole of methanol? (Assume that water is in excess.)

 

[Borek]

Start with the reaction equation.

 

[Blueshift5]

To find the moles of methyl iodide required, we can use the stoichiometry of the reaction to calculate the theoretical yield of methanol. Since the percent yield is given as 50%, we can assume that only half of the theoretical yield will be obtained in the actual reaction.

The balanced chemical equation for the reaction is: CH3I + H2O → CH3OH + HI

From the equation, we can see that for every 1 mole of CH3I reacted, 1 mole of CH3OH is produced. So, if the theoretical yield of CH3OH is 0.1 mole, then the moles of CH3I required would also be 0.1 mole.

However, since the percent yield is 50%, we need to divide the moles of CH3I by 0.5 to get the actual amount required. Therefore, the moles of CH3I needed to prepare 0.1 mole of methanol is 0.1 mole / 0.5 = 0.2 moles.

In summary, 0.2 moles of methyl iodide will be required to prepare 0.1 mole of methanol with a 50% yield. It is important to note that this calculation assumes that water is in excess, meaning that there is more than enough water present in the reaction to react with all of the methyl iodide. If this is not the case, the amount of methyl iodide needed may be different.