D.) It depends cuz no yeild is 100%..I mean side reactions also occur in most of the reactions. So mass of the reactant is not equal to the mass of the product. Hope it helps
Signs that a chemical reaction is occurring are: 1. change in color 2. change in odor 3. change in pH, as in changes from acid to base or base to acid
% yield = 74.35
<h3>Further explanation</h3>
Given
12 dm³ ethene = 12 L
18.4 g ethanol(actual)
Required
The percentage yield
Solution
Reaction
C₂H₄(g) + H₂O(g) ⇒ C₂H₅OH(g)
Assume at STP, 1 mol gas = 22.4 L
mol ethene :
= 12 L : 22.4 L
= 0.538
From equation, mol ratio C₂H₅OH : C₂H₄ = 1 : 1, so mol C₂H₅OH = 0.538
Mass Ethanol (MW=46 g/mol) ⇒ theoretical
= 0.538 x 46
= 24.748 g
% yield = (actual/theoretical) x 100%
% yield = (18.4/24.748) x 100%
% yield = 74.35
Explanation:
The reaction equation will be as follows.
![CO_{2}(aq) + H_{2}O \rightleftharpoons H^{+}(aq) + HCO^{-}_{3}(aq)](https://tex.z-dn.net/?f=CO_%7B2%7D%28aq%29%20%2B%20H_%7B2%7DO%20%5Crightleftharpoons%20H%5E%7B%2B%7D%28aq%29%20%2B%20HCO%5E%7B-%7D_%7B3%7D%28aq%29)
Calculate the amount of
dissolved as follows.
![CO_{2}(aq) = K_{CO_{2}} \times P_{CO_{2}}](https://tex.z-dn.net/?f=CO_%7B2%7D%28aq%29%20%3D%20K_%7BCO_%7B2%7D%7D%20%5Ctimes%20P_%7BCO_%7B2%7D%7D)
It is given that
= 0.032 M/atm and
=
atm.
Hence,
will be calculated as follows.
=
= ![0.032 M/atm \times 1.9 \times 10^{-4}atm](https://tex.z-dn.net/?f=0.032%20M%2Fatm%20%5Ctimes%201.9%20%5Ctimes%2010%5E%7B-4%7Datm)
= ![0.0608 \times 10^{-4}](https://tex.z-dn.net/?f=0.0608%20%5Ctimes%2010%5E%7B-4%7D)
or, = ![0.608 \times 10^{-5}](https://tex.z-dn.net/?f=0.608%20%5Ctimes%2010%5E%7B-5%7D)
It is given that ![K_{a} = 4.46 \times 10^{-7}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%204.46%20%5Ctimes%2010%5E%7B-7%7D)
As, ![K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5E%7B2%7D%7D%7B%5BCO_%7B2%7D%5D%7D)
= ![2.71 \times 10^{-12}](https://tex.z-dn.net/?f=2.71%20%5Ctimes%2010%5E%7B-12%7D)
= ![1.64 \times 10^{-6}](https://tex.z-dn.net/?f=1.64%20%5Ctimes%2010%5E%7B-6%7D)
Since, we know that pH = ![-log [H^{+}]](https://tex.z-dn.net/?f=-log%20%5BH%5E%7B%2B%7D%5D)
So, pH = ![-log (1.64 \times 10^{-6})](https://tex.z-dn.net/?f=-log%20%281.64%20%5Ctimes%2010%5E%7B-6%7D%29)
= 5.7
Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.
Answer:
10moles of kcl
Explanation:
2
K
C
l
O3 → 2
K
C
l + 3
O
2
Notice that you have a 2
:
3 mole ratio between potassium chlorate and oxygen gas, which means that, regardless of how many moles of the former react, you'll always produce 2/3 times more moles of the latter.
15 mol of O2 * ((2mol of KCLO3)/(3mol of O2))= 15*2/3=10 Mol