Answer:
63.53% yield
Explanation:
The balanced equation for this reaction is 2NaCl + H2O -> 2NaOH +Cl2
First we must find the limiting reactant
From NaCl we can only produce 6.06 grams of Cl2 in <u>theory</u>
From H20 we can only produce 38.995 grams in theory
so we know NaCl is the limiting
% yield is (Actual/Theoretical) x100 so
(3.85/6.06)x100= 63.53% yield
First we have to find Ka1 and Ka2
pKa1 = - log Ka1 so Ka1 = 0.059
pKa2 = - log Ka2 so Ka2 = 6.46 x 10⁻⁵
Looking at the values of equilibrium constants we can see that the first one is really big compared to second one. so, the pH will be affected mainly by the first ionization of the acid.
Oxalic acid is H₂C₂O₄
H₂C₂O₄ ⇄ H⁺ + HC₂O₄⁻
0.0356 M 0 0
0.0356 - x x x
Ka1 =
![\frac{[H^+][HC2O4^-]}{[H2C2O4]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BH%5E%2B%5D%5BHC2O4%5E-%5D%7D%7B%5BH2C2O4%5D%7D%20)
= x² / 0.0356 - x
x = 0.025 M
pH = - log [H⁺] = - log (0.025) = 1.6
Relative formula mass C₅H₁₁ = 71
Now divide the molar mass by the RFM = 142.32 / 71 = 2
Now C₍₅ₓ₂₎H₍₁₁ₓ₂) = C₁₀H₂₂
Hope that helps
Answer:I believe your answer would be option B) dilute. Hope this helps.
Explanation:
D 4 grams
i need to write a couple more character for explanation so there