2KClO3 --> 2KClO2 + O2
12 6 (moles)
The ratio of KClO3 and O2 is 2:1. This means 2 moles of KClO3 can create 1 mole of O2. So 12 moles of KClO3 will create 6 moles of O2.
Answer:
0.3758moles
Explanation:
moles of kcl = mass of kcl/ molar mass of kcl = 28/74.5 = 0.3758moles
Explanation:
We have to calculate 
value.
It is known that at the equivalence point concentration of acid is equal to the concentration of anion formed.
Hence, [HA] = ![[A^{-}]](https://tex.z-dn.net/?f=%5BA%5E%7B-%7D%5D)
Now, relation between and pH is as follows.
pH = ![pK_{a} + log \frac{[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=pK_%7Ba%7D%20%2B%20log%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
Putting the values into the above formula as follows.
pH =
4.23 = 
(as [HA] = )
= 4.23 (as log (1) = 0)
or, = 4
Thus, we can conclude that of given weak acid is 4.
Answer:
Percent yield = 61.58%
Explanation:
C4H9OH + NaBr + H2SO4 --> C4H9Br + NaHSO4 + H2O
From the reaction;
1 mol of C4H9OH reacts with 1 mol of NaBr and 1 mol of H2SO4 to form 1 mol of C4H9Br
We have to obtain the limiting reagent. We do this by calculatig the mols of each reactant in the reaction;
Number of moles = Mass / Molar mass
C4H9OH; 15 / 74g/mol = 0.2027
NaBr; 22.4 / 102.89 = 0.2177
H2SO4; 32.7 / 98 = 0.3336
In this case, the limiting reagent is C4H9OH, it determines the amount of product formed.
From the stoichometry of the reaction; one mole of C4H9OH forms one mole of C4H9Br. This means 0.2027 mol of C4H9Br was formed.
Mass = Number of moles * Molar mass
Mass of C4H9Br formed = 0.2027 * 137
Mass = 27.77g
Percent yield = (Practical yield / theoretical yeield ) * 100%
Percent Yield = 17.1 / 27.77 * 100
Percent yield = 61.58%
