MNaHCO₃: 23+1+12+(48×3) = 84g
mCH₃COOH: 12+(1×3)+12+(16×2)+1 = 60g
.................
84g NaHCO₃ react with 60g CH₃COOH
83g NaHCO₃ react with...........
84g ----- 60g
83g ----- X
X = 59,29g CH₃COOH
We used 70g CH₃COOH, it' too much.
So, acetic acid is excess reagent, and sodium bicarbonate is limiting reagent.
_______________________________
B) Amount of CH3COOH is in excess.
:•)
There are 0.109 moles of NaCl in one tablespoon of salt. This is found by dividing the number of grams you have by the molecular weight.
Answer:
pH=10.97
Explanation:
the solution of methyl amine with methylammonium chloride will make a buffer solution.
The pH of buffer solution can be obtained using Henderson Hassalbalch's equation, which is:
![pOH=pKb+log\frac{[salt]}{[base]}](https://tex.z-dn.net/?f=pOH%3DpKb%2Blog%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bbase%5D%7D)
pH = 14- pOH
Let us calculate pOH

[Salt] = [methylammonium chloride] = 0.10 M (initial)
After adding base
![[salt] = \frac{molarityXvolume}{finalvolume}=\frac{0.1X20}{(20+50)}= 0.0286M](https://tex.z-dn.net/?f=%5Bsalt%5D%20%3D%20%5Cfrac%7BmolarityXvolume%7D%7Bfinalvolume%7D%3D%5Cfrac%7B0.1X20%7D%7B%2820%2B50%29%7D%3D%200.0286M)
[base] = [Methylamine]=0.10
After mixing with salt
![[base]= \frac{molarityXvolume}{finalvolume}=\frac{0.1X50}{(20+50)}= 0.0714M](https://tex.z-dn.net/?f=%5Bbase%5D%3D%20%5Cfrac%7BmolarityXvolume%7D%7Bfinalvolume%7D%3D%5Cfrac%7B0.1X50%7D%7B%2820%2B50%29%7D%3D%200.0714M)
pKb= -log[Kb]= 3.43
Putting values
pOH = ![3.43+log(\frac{[0.0286]}{0.0714}](https://tex.z-dn.net/?f=3.43%2Blog%28%5Cfrac%7B%5B0.0286%5D%7D%7B0.0714%7D)
I just did guessed on the question and got it right. The answer is kinetic energy.