The balanced chemical equation is given as:
2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(l)
We are given the yield of CH3CH2OCH2CH3 and the amount of ethanol to be used for the reaction. These values will be the starting point for the calculations.
Theoretical amount of product produced:
329 g CH3CH2OH ( 1 mol / 46.07 g ) ( 1 mol CH3CH2OCH2CH3 / 2 mol CH3CH2OH ) (74.12 g / mol ) = 264.66 g CH3CH2OCH2CH3
% yield = .775 = actual yield / 264.66
actual yield = 205.11 g CH3CH2OCH2CH3
Ar I did this I think i got it right on edunuty
<u>Answer:</u> The pH of resulting solution is 8.7
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:

Molarity of TRIS acid solution = 0.1 M
Volume of solution = 50 mL
Putting values in above equation, we get:

Molarity of TRIS base solution = 0.2 M
Volume of solution = 60 mL
Putting values in above equation, we get:

Volume of solution = 50 + 60 = 110 mL = 0.11 L (Conversion factor: 1 L = 1000 mL)
- To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
![pH=pK_a+\log(\frac{[salt]}{[acid]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D%29)
![pH=pK_a+\log(\frac{[\text{TRIS base}]}{[\text{TRIS acid}]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5B%5Ctext%7BTRIS%20base%7D%5D%7D%7B%5B%5Ctext%7BTRIS%20acid%7D%5D%7D%29)
We are given:
= negative logarithm of acid dissociation constant of TRIS acid = 8.3
![[\text{TRIS acid}]=\frac{0.005}{0.11}](https://tex.z-dn.net/?f=%5B%5Ctext%7BTRIS%20acid%7D%5D%3D%5Cfrac%7B0.005%7D%7B0.11%7D)
![[\text{TRIS base}]=\frac{0.012}{0.11}](https://tex.z-dn.net/?f=%5B%5Ctext%7BTRIS%20base%7D%5D%3D%5Cfrac%7B0.012%7D%7B0.11%7D)
pH = ?
Putting values in above equation, we get:

Hence, the pH of resulting solution is 8.7
<h2>Answer -
0.73%</h2>
14.6 .2
=
100 x
(Above is proportions I used to get the answer)
x = 0.73% which is the answer.